Fundamental Concepts of Algebra

A term made up of numbers (digits) only is a constant term, whereas a term containing one or more letters (variables) is a variable term.

Constant term = 8

Variable terms = $\bm{x, 6xy, 6 + x, -5xy^2, 15az^2, \frac{32z}{xy}}$ and $\bm{\frac{y^2}{3x}}$

An expression with 1 term is a monomial, with 2 unlike terms is a binomial and with 3 unlike terms is a trinomial.

(i) $2x \div 15 = \frac{2x}{15}$, which has 1 term → Monomial.

(ii) $ax + 9$ has 2 terms → Binomial.

(iii) $3x^2 \times 5x = 15x^3$, which has 1 term → Monomial.

(iv) $5 + 2a - 3b$ has 3 terms → Trinomial.

(v) $2y - \frac{7}{3}z \div x = 2y - \frac{7z}{3x}$, which has 2 terms → Binomial.

(vi) $3p \times q \div z = \frac{3pq}{z}$, which has 1 term → Monomial.

(vii) $12z \div 5x + 4 = \frac{12z}{5x} + 4$, which has 2 terms → Binomial.

(viii) $12 - 5z - 4 = 8 - 5z$, which has 2 terms → Binomial.

(ix) $a^3 - 3ab^2 \times c = a^3 - 3ab^2c$, which has 2 terms → Binomial.

$-3axy = (-3a) \times xy$, so the coefficient of $xy$ is $\bm{-3a}$.

$p^2yz^2 = (p^2y) \times z^2$, so the coefficient of $z^2$ is $\bm{p^2y}$.

$-mn = (-1) \times mn$, so the coefficient of $mn$ is $\bm{-1}$.

$-15p^2 = 15 \times (-p^2)$, so the coefficient of 15 is $\bm{-p^2}$.

The degree of a monomial is the sum of the exponents of its variables.

Degree of $7y = \bm{1}$.

The degree of a monomial is the sum of the exponents of its variables.

Degree of $-x^2y = 2 + 1 = \bm{3}$.

The degree of a monomial is the sum of the exponents of its variables.

Degree of $xy^2z = 1 + 2 + 1 = \bm{4}$.

The degree of a monomial is the sum of the exponents of its variables.

Degree of $-9y^2z^3 = 2 + 3 = \bm{5}$.

The degree of a monomial is the sum of the exponents of its variables.

Degree of $3m^3n^4 = 3 + 4 = \bm{7}$.

The degree of a monomial is the sum of the exponents of its variables.

Degree of $-2p^2q^3r^4 = 2 + 3 + 4 = \bm{9}$.

The degree of a polynomial is the degree of its highest degree term.

Degrees of the terms are 3, (2 + 2) = 4 and 1. The greatest is 4.

Hence, the degree of the polynomial = 4.

The degree of a polynomial is the degree of its highest degree term.

Degrees of the terms are (3 + 2) = 5, (2 + 5) = 7 and (4 + 4) = 8. The greatest is 8.

Hence, the degree of the polynomial = 8.

The degree of a polynomial is the degree of its highest degree term.

Degrees of the terms are (1 + 6) = 7 and (3 + 1) = 4. The greatest is 7.

Hence, the degree of the polynomial = 7.

The degree of a polynomial is the degree of its highest degree term.

Degrees of the terms are 0, (2 + 1) = 3 and 2. The greatest is 3.

Hence, the degree of the polynomial = 3.

The degree of a polynomial is the degree of its highest degree term.

Degrees of the terms are 1 and 0. The greatest is 1.

Hence, the degree of the polynomial = 1.

The degree of a polynomial is the degree of its highest degree term.

Degrees of the terms are (2 + 1) = 3 and (1 + 3) = 4. The greatest is 4.

Hence, the degree of the polynomial = 4.

Terms having the same literal (variable) part are like terms.

The groups of like terms are $\bm{9x^2, -3x^2}$ and $\bm{x^2}$; $\bm{xy}$ and $\bm{-2xy}$.

Terms having the same literal (variable) part are like terms.

The groups of like terms are $\bm{ab}$ and $\bm{-3ab}$; $\bm{-a^2b, 5a^2b}$ and $\bm{-8a^2b}$.

Terms having the same literal (variable) part are like terms.

The groups of like terms are $\bm{7p, -2p}$ and $\bm{3p}$; $\bm{8pq}$ and $\bm{-5pq}$.

(i) $y = 1 \times y$, so its numerical coefficient is $\bm{1}$.

(ii) $-y = (-1) \times y$, so its numerical coefficient is $\bm{-1}$.

(iii) The numerical coefficient of $2x^2y$ is $\bm{2}$.

(iv) The numerical coefficient of $-8xy^3$ is $\bm{-8}$.

(v) The numerical coefficient of $3py^2$ is $\bm{3}$.

(vi) The numerical coefficient of $-9a^2b^3$ is $\bm{-9}$.

We have $-5x^3y^2z^4$.

(i) $-5x^3y^2z^4 = (-5x^3y^2z^2) \times z^2$, so the coefficient of $z^2$ is $\bm{-5x^3y^2z^2}$.

(ii) $-5x^3y^2z^4 = (-5x^3z^4) \times y^2$, so the coefficient of $y^2$ is $\bm{-5x^3z^4}$.

(iii) $-5x^3y^2z^4 = (-5x^3yz^2) \times yz^2$, so the coefficient of $yz^2$ is $\bm{-5x^3yz^2}$.

(iv) $-5x^3y^2z^4 = (x^3y)(-5yz^4)$, so the coefficient of $x^3y$ is $\bm{-5yz^4}$.

(v) $-5x^3y^2z^4 = (-xy^2)(5x^2z^4)$, so the coefficient of $-xy^2$ is $\bm{5x^2z^4}$.

(vi) $-5x^3y^2z^4 = (-5xy^2z)(x^2z^3)$, so the coefficient of $-5xy^2z$ is $\bm{x^2z^3}$.

Degree of $-5x^3y^2z^4 = 3 + 2 + 4 = 9$.

Hence, the degree of the given algebraic expression = 9.

An algebraic expression is a polynomial only if the exponent of each variable in every term is a whole number.

(i) In $\frac{8}{5}x^3 + 7$, the exponent of $x$ is 3 which is a whole number → It is a polynomial.

(ii) $3\sqrt{x} = 3x^{\frac{1}{2}}$ has a fractional exponent → Not a polynomial.

(iii) $\frac{8}{x^3} = 8x^{-3}$ has a negative exponent → Not a polynomial.

(iv) In $3^x$, the exponent itself is a variable → Not a polynomial.

(v) $\frac{18}{x^3} = 18x^{-3} \text{ and } \frac{7}{y^2} = 7y^{-2}$ have negative exponents → Not a polynomial.

Hence, only expression (i) is a polynomial.

Sum of $x$ and $y = x + y$.

Adding 15 to it, the required expression is $\bm{(x + y) + 15}$.

Sum of $2m$ and $3y = 2m + 3y$.

Subtracting 23 from it, the required expression is $\bm{(2m + 3y) - 23}$.

Sum of $a$ and $b = a + b$ and their product = $ab$.

So, the required expression is $\bm{(a + b) + ab}$.

Product of $x$ and $y = xy$ and their sum = $x + y$.

So, the required expression is $\bm{xy - (x + y)}$.

$x \div y = \frac{x}{y}$ and $y \div x = \frac{y}{x}$

So, the required expression is $\bm{\frac{x}{y} + \frac{y}{x} = (x \div y) + (y \div x)}$.

(i) $8x + 5x = (8 + 5)x = \bm{13x}$.

(ii) $8x - 5x = (8 - 5)x = \bm{3x}$.

(iii) $6xy^2 + 9xy^2 = (6 + 9)xy^2 = \bm{15xy^2}$.

(iv) $6xy^2 - 9xy^2 = (6 - 9)xy^2 = \bm{-3xy^2}$.

(v) $8a + 6a + 5b = (8 + 6)a + 5b = \bm{14a + 5b}$.

(vi) $5 + 7xy + 6 + 3xy = (5 + 6) + (7 + 3)xy = \bm{11 + 10xy}$.

(vii) $4a + 3b - 7a + 4b = (4 - 7)a + (3 + 4)b = \bm{-3a + 7b}$.

(viii) $-15x + 13x + 8 = (-15 + 13)x + 8 = \bm{-2x + 8}$.

(ix) $6x^2y + 13xy^2 - 4x^2y + 2xy^2 = (6 - 4)x^2y + (13 + 2)xy^2 = \bm{2x^2y + 15xy^2}$.

(x) $16x^2 - 9x^2 = \bm{7x^2}$ and $25xy^2 - 17xy^2 = \bm{8xy^2}$.

Solving,

$$\begin{aligned}\Rightarrow -9x + 3x + 4x \\[1em] \Rightarrow (-9 + 3 + 4)x \\[1em] \Rightarrow -2x\end{aligned}$$

Hence, the required sum = $\bm{-2x}$.

Solving,

$$\begin{aligned}\Rightarrow 23y^2 + 8y^2 + (-12y^2) \\[1em] \Rightarrow (23 + 8 - 12)y^2 \\[1em] \Rightarrow 19y^2\end{aligned}$$

Hence, the required sum = $\bm{19y^2}$.

Solving,

$$\begin{aligned}\Rightarrow 18pq + (-15pq) + 3pq \\[1em] \Rightarrow (18 - 15 + 3)pq \\[1em] \Rightarrow 6pq\end{aligned}$$

Hence, the required sum = $\bm{6pq}$.

Solving,

$$\begin{aligned}\Rightarrow 3m + 12m - 5m \\[1em] \Rightarrow (3 + 12 - 5)m \\[1em] \Rightarrow 10m\end{aligned}$$

Hence, the simplified expression = $\bm{10m}$.

Solving,

$$\begin{aligned}\Rightarrow 7n^2 - 9n^2 + 3n^2 \\[1em] \Rightarrow (7 - 9 + 3)n^2 \\[1em] \Rightarrow n^2\end{aligned}$$

Hence, the simplified expression = $\bm{n^2}$.

Solving,

$$\begin{aligned}\Rightarrow 25zy - 8zy - 6zy \\[1em] \Rightarrow (25 - 8 - 6)zy \\[1em] \Rightarrow 11yz\end{aligned}$$

Hence, the simplified expression = $\bm{11yz}$.

Solving,

$$\begin{aligned}\Rightarrow -5ax^2 + 7ax^2 - 12ax^2 \\[1em] \Rightarrow (-5 + 7 - 12)ax^2 \\[1em] \Rightarrow -10ax^2\end{aligned}$$

Hence, the simplified expression = $\bm{-10ax^2}$.

Solving,

$$\begin{aligned}\Rightarrow -16am + 4mx + 4am - 15mx + 5am \\[1em] \Rightarrow (-16 + 4 + 5)am + (4 - 15)mx \\[1em] \Rightarrow -7am - 11mx\end{aligned}$$

Hence, the simplified expression = $\bm{-7am - 11mx}$.

Solving,

$$\begin{array}{rrr} & a & + b \\ + & 2a & + 3b \\ \hline & 3a & + 4b \\ \end{array}$$

Hence, the required sum = $\bm{3a + 4b}$.

Solving,

$$\begin{array}{rrr} & 2x & + y \\ + & 3x & - 4y \\ \hline & 5x & - 3y \\ \end{array}$$

Hence, the required sum = $\bm{5x - 3y}$.

Solving,

$$\begin{array}{rrr} & -3a & + 2b \\ + & 3a & + b \\ \hline & & 3b \\ \end{array}$$

Hence, the required sum = $\bm{3b}$.

Solving,

$$\begin{array}{rrr} & 4 & + x \\ & 5 & - 2x \\ + & & 6x \\ \hline & 9 & + 5x \\ \end{array}$$

Hence, the required sum = $\bm{9 + 5x}$.

Solving,

$$\begin{array}{rrrr} & 3x & + 8y & + 7z \\ & -2x & + 6y & + 4z \\ + & -4x & + 3y & + 6z \\ \hline & -3x & + 17y & + 17z \\ \end{array}$$

Hence, the required sum = $\bm{-3x + 17y + 17z}$.

Solving,

$$\begin{array}{rrrr} & 3a & + 5b & + 2c \\ & 2a & + 3b & - c \\ + & a & + b & + c \\ \hline & 6a & + 9b & + 2c \\ \end{array}$$

Hence, the required sum = $\bm{6a + 9b + 2c}$.

Solving,

$$\begin{array}{rrrr} & 4x^2 & + 8xy & - 2y^2 \\ + & x^2 & + 8xy & - 5y^2 \\ \hline & 5x^2 & + 16xy & - 7y^2 \\ \end{array}$$

Hence, the required sum = $\bm{5x^2 + 16xy - 7y^2}$.

Solving,

$$\begin{array}{rrrr} & 9x^2 & - 6x & + 7 \\ & & -4x & + 5 \\ + & -3x^2 & & + 6 \\ \hline & 6x^2 & - 10x & + 18 \\ \end{array}$$

Hence, the required sum = $\bm{6x^2 - 10x + 18}$.

Solving,

$$\begin{array}{rrrr} & 5x^2 & - 2xy & + 3y^2 \\ & -2x^2 & + 5xy & + 9y^2 \\ + & 3x^2 & - xy & - 4y^2 \\ \hline & 6x^2 & + 2xy & + 8y^2 \\ \end{array}$$

Hence, the required sum = $\bm{6x^2 + 2xy + 8y^2}$.

Solving,

$$\begin{array}{rrrrrrr} & a^2 & + b^2 & & + 2ab & & \\ & & 2b^2 & + c^2 & & + 2bc & \\ + & -a^2 & & + 4c^2 & & & + 2ac \\ \hline & & 3b^2 & + 5c^2 & + 2ab & + 2bc & + 2ac \\ \end{array}$$

Hence, the required sum = $\bm{3b^2 + 5c^2 + 2ab + 2bc + 2ac}$.

Solving,

$$\begin{array}{rrrr} & 9ax & - 6bx & + 8 \\ & 4ax & + 8bx & - 7 \\ + & -6ax & - 4bx & - 3 \\ \hline & 7ax & - 2bx & - 2 \\ \end{array}$$

Hence, the required sum = $\bm{7ax - 2bx - 2}$.

Solving,

$$\begin{array}{rrrr} & abc & + 2ab & + 3ac \\ & 2abc & - 4ab & + 4ac \\ + & -3abc & + 2ab & - 6ac \\ \hline & & & ac \\ \end{array}$$

Hence, the required sum = $\bm{ac}$.

Solving,

$$\begin{array}{rrrr} & 4a^2 & + 5b^2 & - 6ab \\ & & & 3ab \\ & 6a^2 & - 2b^2 & \\ + & & 4b^2 & - 5ab \\ \hline & 10a^2 & + 7b^2 & - 8ab \\ \end{array}$$

Hence, the required sum = $\bm{10a^2 + 7b^2 - 8ab}$.

Solving,

$$\begin{array}{rrrr} & x^2 & + x & - 2 \\ & -3x^2 & + 2x & + 5 \\ + & 2x^2 & - 5x & + 7 \\ \hline & & -2x & + 10 \\ \end{array}$$

Hence, the required sum = $\bm{-2x + 10}$.

Solving,

$$\begin{array}{rrrrr} & 4x^3 & + 2x^2 & - x & + 1 \\ & 2x^3 & - 5x^2 & - 3x & + 6 \\ & & x^2 & & + 8 \\ + & 5x^3 & & - 7x & \\ \hline & 11x^3 & - 2x^2 & - 11x & + 15 \\ \end{array}$$

Hence, the required sum = $\bm{11x^3 - 2x^2 - 11x + 15}$.

Since the given terms are unlike terms, their sum cannot be simplified further.

Sum = $\bm{x + 3y}$.

Since the given terms are unlike terms, their sum cannot be simplified further.

Sum = $\bm{-2a + 5}$.

Since the given terms are unlike terms, their sum cannot be simplified further.

Sum = $\bm{-4x^2 + 7x}$.

Since the given terms are unlike terms, their sum cannot be simplified further.

Sum = $\bm{4a - 7b}$.

Since the given terms are unlike terms, their sum cannot be simplified further.

Sum = $\bm{x^3 + 3x^2y + 2y^2}$ .

Since the given terms are unlike terms, their sum cannot be simplified further.

Sum = $\bm{11 - by}$.

Perimeter of a triangle = sum of its three sides

Solving,

$$\begin{aligned}\Rightarrow (2x + 3y) + (x + 5y) + (7x - 2y) \\[1em] \Rightarrow (2 + 1 + 7)x + (3 + 5 - 2)y \\[1em] \Rightarrow 10x + 6y\end{aligned}$$

Hence, the perimeter of the triangle = $\bm{10x + 6y}$.

Perimeter of a rectangle = 2 × (sum of two adjacent sides)

Solving,

$$\begin{aligned}\Rightarrow 2 \times [(6a + 9b) + (8a - 4b)] \\[1em] \Rightarrow 2 \times (14a + 5b) \\[1em] \Rightarrow 28a + 10b\end{aligned}$$

Hence, the perimeter of the rectangle = $\bm{28a + 10b}$.

Solving,

$$\begin{aligned}\Rightarrow (2a + b) - (a + b) \\[1em] \Rightarrow 2a + b - a - b \\[1em] \Rightarrow 2a - a + b - b \\[1em] \Rightarrow a.\end{aligned}$$

Hence, the required difference = $\bm{a}$.

Solving,

$$\begin{aligned}\Rightarrow (-2b + 2c) - (b + 3c) \\[1em] \Rightarrow -2b + 2c - b - 3c \\[1em] \Rightarrow -3b - c\end{aligned}$$

Hence, the required difference = $\bm{-3b - c}$.

Solving,

$$\begin{aligned}\Rightarrow (5a + b) - (-6b + 2a) \\[1em] \Rightarrow 5a + b + 6b - 2a \\[1em] \Rightarrow 3a + 7b\end{aligned}$$

Hence, the required difference = $\bm{3a + 7b}$.

Solving,

$$\begin{aligned}\Rightarrow (a^3 - 1 + a) - (3a - 2a^2) \\[1em] \Rightarrow a^3 - 1 + a - 3a + 2a^2 \\[1em] \Rightarrow a^3 + 2a^2 - 2a - 1\end{aligned}$$

Hence, the required difference = $\bm{a^3 + 2a^2 - 2a - 1}$.

Solving,

$$\begin{aligned}\Rightarrow (p + 2) - 1 \\[1em] \Rightarrow p + 1\end{aligned}$$

Hence, the required difference = $\bm{p + 1}$.

Solving,

$$\begin{aligned}\Rightarrow (x + 2y + z) - (-x - y - 3z) \\[1em] \Rightarrow x + 2y + z + x + y + 3z \\[1em] \Rightarrow 2x + 3y + 4z\end{aligned}$$

Hence, the required difference = $\bm{2x + 3y + 4z}$.

Solving,

$$\begin{aligned}\Rightarrow (3a^2 - 8ab - 2b^2) - (3a^2 - 4ab + 6b^2) \\[1em] \Rightarrow 3a^2 - 8ab - 2b^2 - 3a^2 + 4ab - 6b^2 \\[1em] \Rightarrow -4ab - 8b^2\end{aligned}$$

Hence, the required difference = $\bm{-4ab - 8b^2}$.

Solving,

$$\begin{aligned}\Rightarrow (4pq - 6p^2 - 2q^2) - 9p^2 \\[1em] \Rightarrow 4pq - 6p^2 - 2q^2 - 9p^2 \\[1em] \Rightarrow 4pq - 15p^2 - 2q^2\end{aligned}$$

Hence, the required difference = $\bm{4pq - 15p^2 - 2q^2}$.

Solving,

$$\begin{aligned}\Rightarrow 10abc - (2a^2 + 2abc - 4b^2) \\[1em] \Rightarrow 10abc - 2a^2 - 2abc + 4b^2 \\[1em] \Rightarrow -2a^2 + 8abc + 4b^2\end{aligned}$$

Hence, the required difference = $\bm{-2a^2 + 8abc + 4b^2}$.

Solving,

$$\begin{aligned}\Rightarrow (a^2 + ab + c^2) - (a^2 - d^2) \\[1em] \Rightarrow a^2 + ab + c^2 - a^2 + d^2 \\[1em] \Rightarrow ab + c^2 + d^2\end{aligned}$$

Hence, the required difference = $\bm{ab + c^2 + d^2}$.

Solving,

$$\begin{aligned}\Rightarrow (8 - x) - 4x \\[1em] \Rightarrow 8 - x - 4x \\[1em] \Rightarrow 8 - 5x\end{aligned}$$

Hence, the required difference = $\bm{8 - 5x}$.

Solving,

$$\begin{aligned}\Rightarrow (c + 3d) - (-8c) \\[1em] \Rightarrow c + 3d + 8c \\[1em] \Rightarrow 9c + 3d\end{aligned}$$

Hence, the required difference = $\bm{9c + 3d}$.

Solving,

$$\begin{aligned}\Rightarrow (b + 6c) - (-5a - 2b) \\[1em] \Rightarrow b + 6c + 5a + 2b \\[1em] \Rightarrow 5a + 3b + 6c\end{aligned}$$

Hence, the required difference = $\bm{5a + 3b + 6c}$.

Solving,

$$\begin{aligned}\Rightarrow (3p^2 - 8p) - (4p + p^2) \\[1em] \Rightarrow 3p^2 - 8p - 4p - p^2 \\[1em] \Rightarrow 2p^2 - 12p\end{aligned}$$

Hence, the required difference = $\bm{2p^2 - 12p}$.

Solving,

$$\begin{aligned}\Rightarrow (4a - b - 2c) - (5a - 3b + 2c) \\[1em] \Rightarrow 4a - b - 2c - 5a + 3b - 2c \\[1em] \Rightarrow -a + 2b - 4c\end{aligned}$$

Hence, the required difference = $\bm{-a + 2b - 4c}$.

Solving,

$$\begin{aligned}\Rightarrow (xy - yz + xz) - (-xy + yz - zx) \\[1em] \Rightarrow xy - yz + xz + xy - yz + zx \\[1em] \Rightarrow 2xy - 2yz + 2xz \\[1em] \Rightarrow 2(xy - yz + xz)\end{aligned}$$

Hence, the required difference = $\bm{2(xy - yz + xz)}$.

Solving,

$$\begin{aligned}\Rightarrow (3x^2 - 5xy + 3y^2) - (2x^2 - 7xy - y^2) \\[1em] \Rightarrow 3x^2 - 5xy + 3y^2 - 2x^2 + 7xy + y^2 \\[1em] \Rightarrow x^2 + 2xy + 4y^2\end{aligned}$$

Hence, the required difference = $\bm{x^2 + 2xy + 4y^2}$.

Solving,

$$\begin{aligned}\Rightarrow (2b^2 - a^2 + 2ab) - (a^2 - 3ab - 6b^2) \\[1em] \Rightarrow 2b^2 - a^2 + 2ab - a^2 + 3ab + 6b^2 \\[1em] \Rightarrow -2a^2 + 5ab + 8b^2\end{aligned}$$

Hence, the required difference = $\bm{-2a^2 + 5ab + 8b^2}$.

Solving,

$$\begin{aligned}\Rightarrow (-3y^2 + 5xy^2 - 7x^2 - 9x^2y) - (4x^2 - 5x^2y + y^2) \\[1em] \Rightarrow -3y^2 + 5xy^2 - 7x^2 - 9x^2y - 4x^2 + 5x^2y - y^2 \\[1em] \Rightarrow (-7 - 4)x^2 + (-9 + 5)x^2y + 5xy^2 + (-3 - 1)y^2 \\[1em] \Rightarrow -11x^2 - 4x^2y + 5xy^2 - 4y^2\end{aligned}$$

Hence, the required difference = $\bm{-11x^2 - 4x^2y + 5xy^2 - 4y^2}$.

Solving,

$$\begin{aligned}\Rightarrow (3m^3 + 4) - (6m^3 + 4m^2 + 7m - 3) \\[1em] \Rightarrow 3m^3 + 4 - 6m^3 - 4m^2 - 7m + 3 \\[1em] \Rightarrow -3m^3 - 4m^2 - 7m + 7\end{aligned}$$

Hence, the required difference = $\bm{-3m^3 - 4m^2 - 7m + 7}$.

Sum of $4a^2 + 3 - 8a$ and $9a - 7$,

$$\begin{aligned}\Rightarrow (4a^2 + 3 - 8a) + (9a - 7) \\[1em] \Rightarrow 4a^2 + (-8 + 9)a + (3 - 7) \\[1em] \Rightarrow 4a^2 + a - 4\end{aligned}$$

Now, subtracting $-5a^2 - 3a + 1$ from this sum,

$$\begin{aligned}\Rightarrow (4a^2 + a - 4) - (-5a^2 - 3a + 1) \\[1em] \Rightarrow 4a^2 + a - 4 + 5a^2 + 3a - 1 \\[1em] \Rightarrow 9a^2 + 4a - 5\end{aligned}$$

Hence, the required difference = $\bm{9a^2 + 4a - 5}$.

Solving,

$$\begin{aligned}\Rightarrow (8x^3 - 6x^2 + 9x - 10) - (4x^3 + 2x^2 + 7x - 3) \\[1em] \Rightarrow 8x^3 - 6x^2 + 9x - 10 - 4x^3 - 2x^2 - 7x + 3 \\[1em] \Rightarrow (8 - 4)x^3 + (-6 - 2)x^2 + (9 - 7)x + (-10 + 3) \\[1em] \Rightarrow 4x^3 - 8x^2 + 2x - 7\end{aligned}$$

Hence, the required excess = $\bm{4x^3 - 8x^2 + 2x - 7}$.

Solving,

$$\begin{aligned}\Rightarrow (a^2 - a - a^3 + 1) - (2a^3 + 5a - a^2 - 6) \\[1em] \Rightarrow a^2 - a - a^3 + 1 - 2a^3 - 5a + a^2 + 6 \\[1em] \Rightarrow (-1 - 2)a^3 + (1 + 1)a^2 + (-1 - 5)a + (1 + 6) \\[1em] \Rightarrow -3a^3 + 2a^2 - 6a + 7\end{aligned}$$

Hence, the required expression = $\bm{-3a^3 + 2a^2 - 6a + 7}$.

Solving,

$$\begin{aligned}\Rightarrow (a^2 + b^2 + 2ab) - (-4ab + 2b^2) \\[1em] \Rightarrow a^2 + b^2 + 2ab + 4ab - 2b^2 \\[1em] \Rightarrow a^2 + (1 - 2)b^2 + (2 + 4)ab \\[1em] \Rightarrow a^2 - b^2 + 6ab\end{aligned}$$

Hence, the required expression = $\bm{a^2 - b^2 + 6ab}$.

Solving,

$$\begin{aligned}\Rightarrow (4m^2 + 4n^2 + 4p^2) - (m^2 + 3n^2 - 5p^2) \\[1em] \Rightarrow 4m^2 + 4n^2 + 4p^2 - m^2 - 3n^2 + 5p^2 \\[1em] \Rightarrow (4 - 1)m^2 + (4 - 3)n^2 + (4 + 5)p^2 \\[1em] \Rightarrow 3m^2 + n^2 + 9p^2\end{aligned}$$

Hence, the required excess = $\bm{3m^2 + n^2 + 9p^2}$.

Solving,

$$\begin{aligned}\Rightarrow (4x^3 - 3x^2y - 7xy^2 + 2y^3) - (3x^3 - 2x^2y + xy^2 - y^3) \\[1em] \Rightarrow 4x^3 - 3x^2y - 7xy^2 + 2y^3 - 3x^3 + 2x^2y - xy^2 + y^3 \\[1em] \Rightarrow (4 - 3)x^3 + (-3 + 2)x^2y + (-7 - 1)xy^2 + (2 + 1)y^3 \\[1em] \Rightarrow x^3 - x^2y - 8xy^2 + 3y^3\end{aligned}$$

Hence, the required difference = $\bm{x^3 - x^2y - 8xy^2 + 3y^3}$.

Sum of $5a^2 - 9a + 3$ and $2a - a^2 - 1$,

$$\begin{aligned}\Rightarrow (5a^2 - 9a + 3) + (2a - a^2 - 1) \\[1em] \Rightarrow (5 - 1)a^2 + (-9 + 2)a + (3 - 1) \\[1em] \Rightarrow 4a^2 - 7a + 2\end{aligned}$$

Sum of $3a^2 - 2a + 5$ and $a^2 - 5a - 7$,

$$\begin{aligned}\Rightarrow (3a^2 - 2a + 5) + (a^2 - 5a - 7) \\[1em] \Rightarrow (3 + 1)a^2 + (-2 - 5)a + (5 - 7) \\[1em] \Rightarrow 4a^2 - 7a - 2\end{aligned}$$

Now, subtracting the second sum from the first sum,

$$\begin{aligned}\Rightarrow (4a^2 - 7a + 2) - (4a^2 - 7a - 2) \\[1em] \Rightarrow 4a^2 - 7a + 2 - 4a^2 + 7a + 2 \\[1em] \Rightarrow 4\end{aligned}$$

Hence, the required difference = $\bm{4}$.

Let the length and the breadth of the rectangle be $l$ and $b$.

Perimeter = $2(l + b)$

$$\begin{aligned}\Rightarrow l + b = \frac{\text{Perimeter}}{2} \\[1em] \Rightarrow l + b = \frac{28x^3 + 16x^2 + 8x + 4}{2} \\[1em] \Rightarrow l + b = 14x^3 + 8x^2 + 4x + 2\end{aligned}$$

Other side = $(l + b)$ − (given side)

$$\begin{aligned}\Rightarrow (14x^3 + 8x^2 + 4x + 2) - (8x^2 + 4x) \\[1em] \Rightarrow 14x^3 + 8x^2 + 4x + 2 - 8x^2 - 4x \\[1em] \Rightarrow 14x^3 + 2\end{aligned}$$

Hence, the other side of the rectangle = $\bm{14x^3 + 2}$.

Sum of the two given sides,

$$\begin{aligned}\Rightarrow (3a^2 + 5a + 1) + (a^2 + 10a - 6) \\[1em] \Rightarrow (3 + 1)a^2 + (5 + 10)a + (1 - 6) \\[1em] \Rightarrow 4a^2 + 15a - 5\end{aligned}$$

Third side = Perimeter − (sum of the two given sides)

$$\begin{aligned}\Rightarrow (14a^2 + 20a + 13) - (4a^2 + 15a - 5) \\[1em] \Rightarrow 14a^2 + 20a + 13 - 4a^2 - 15a + 5 \\[1em] \Rightarrow 10a^2 + 5a + 18\end{aligned}$$

Hence, the third side = $\bm{10a^2 + 5a + 18}$.

Given, $x = 4a^2 + b^2 - 6ab, y = 3b^2 - 2a^2 + 8ab$ and $z = 6a^2 + 8b^2 - 6ab$.

(i) $x + y + z$

$$\begin{aligned}\Rightarrow (4a^2 + b^2 - 6ab) + (3b^2 - 2a^2 + 8ab) + (6a^2 + 8b^2 - 6ab) \\[1em] \Rightarrow (4 - 2 + 6)a^2 + (1 + 3 + 8)b^2 + (-6 + 8 - 6)ab \\[1em] \Rightarrow 8a^2 + 12b^2 - 4ab\end{aligned}$$

Hence, $\bm{x + y + z = 8a^2 + 12b^2 - 4ab}$.

(ii) $x - y - z$

$$\begin{aligned}\Rightarrow (4a^2 + b^2 - 6ab) - (3b^2 - 2a^2 + 8ab) - (6a^2 + 8b^2 - 6ab) \\[1em] \Rightarrow 4a^2 + b^2 - 6ab - 3b^2 + 2a^2 - 8ab - 6a^2 - 8b^2 + 6ab \\[1em] \Rightarrow (4 + 2 - 6)a^2 + (1 - 3 - 8)b^2 + (-6 - 8 + 6)ab \\[1em] \Rightarrow -10b^2 - 8ab\end{aligned}$$

Hence, $\bm{x - y - z = -10b^2 - 8ab}$.

Given, $m = 9x^2 - 4xy + 5y^2$ and $n = -3x^2 + 2xy - y^2$.

(i) $2m - n$

$$\begin{aligned}\Rightarrow 2(9x^2 - 4xy + 5y^2) - (-3x^2 + 2xy - y^2) \\[1em] \Rightarrow 18x^2 - 8xy + 10y^2 + 3x^2 - 2xy + y^2 \\[1em] \Rightarrow 21x^2 - 10xy + 11y^2\end{aligned}$$

Hence, $\bm{2m - n = 21x^2 - 10xy + 11y^2}$.

(ii) $m + 2n$

$$\begin{aligned}\Rightarrow (9x^2 - 4xy + 5y^2) + 2(-3x^2 + 2xy - y^2) \\[1em] \Rightarrow 9x^2 - 4xy + 5y^2 - 6x^2 + 4xy - 2y^2 \\[1em] \Rightarrow 3x^2 + 3y^2\end{aligned}$$

Hence, $\bm{m + 2n = 3x^2 + 3y^2}$.

(iii) $m - 3n$

$$\begin{aligned}\Rightarrow (9x^2 - 4xy + 5y^2) - 3(-3x^2 + 2xy - y^2) \\[1em] \Rightarrow 9x^2 - 4xy + 5y^2 + 9x^2 - 6xy + 3y^2 \\[1em] \Rightarrow 18x^2 - 10xy + 8y^2\end{aligned}$$

Hence, $\bm{m - 3n = 18x^2 - 10xy + 8y^2}$.

Solving,

$$\begin{aligned}\Rightarrow 3x + 5(2x + 6) - 7x \\[1em] \Rightarrow 3x + 10x + 30 - 7x \\[1em] \Rightarrow 6x + 30\end{aligned}$$

Hence, the simplified expression = $\bm{6x + 30}$.

Solving,

$$\begin{aligned}\Rightarrow 3(4y - 10) + 2(y - 1) \\[1em] \Rightarrow 12y - 30 + 2y - 2 \\[1em] \Rightarrow 14y - 32\end{aligned}$$

Hence, the simplified expression = $\bm{14y - 32}$.

Solving,

$$\begin{aligned}\Rightarrow -(7 + 6x) - 7(x + 2) \\[1em] \Rightarrow -7 - 6x - 7x - 14 \\[1em] \Rightarrow -21 - 13x\end{aligned}$$

Hence, the simplified expression = $\bm{-21 - 13x}$.

Solving,

$$\begin{aligned}\Rightarrow x - (x - y) - y - (y - x) \\[1em] \Rightarrow x - x + y - y - y + x \\[1em] \Rightarrow x - y\end{aligned}$$

Hence, the simplified expression = $\bm{x - y}$.

Solving,

$$\begin{aligned}\Rightarrow 4x + 7y - [5y - 8] - 2x \\[1em] \Rightarrow 4x + 7y - 5y + 8 - 2x \\[1em] \Rightarrow 2x + 2y + 8\end{aligned}$$

Hence, the simplified expression = $\bm{2x + 2y + 8}$.

Solving,

$$\begin{aligned}\Rightarrow -2m + 5 + 4(m - 3) \\[1em] \Rightarrow -2m + 5 + 4m - 12 \\[1em] \Rightarrow 2m - 7\end{aligned}$$

Hence, the simplified expression = $\bm{2m - 7}$.

Solving,

$$\begin{aligned}\Rightarrow 2x - y + 5 - (x - y) \\[1em] \Rightarrow 2x - y + 5 - x + y \\[1em] \Rightarrow x + 5\end{aligned}$$

Hence, the simplified expression = $\bm{x + 5}$.

Solving,

$$\begin{aligned}\Rightarrow 2(x - y) - (x - 8) \\[1em] \Rightarrow 2x - 2y - x + 8 \\[1em] \Rightarrow x - 2y + 8\end{aligned}$$

Hence, the simplified expression = $\bm{x - 2y + 8}$.

Solving,

$$\begin{aligned}\Rightarrow 4(3x - 8) - 3(5x + 3) - 2(6x - 8) \\[1em] \Rightarrow 12x - 32 - 15x - 9 - 12x + 16 \\[1em] \Rightarrow (12 - 15 - 12)x + (-32 - 9 + 16) \\[1em] \Rightarrow -15x - 25\end{aligned}$$

Hence, the simplified expression = $\bm{-15x - 25}$.

Solving,

$$\begin{aligned}\Rightarrow 5(x - 4) - 3(x - 4) + 7(x - 4) \\[1em] \Rightarrow (5 - 3 + 7)(x - 4) \\[1em] \Rightarrow 9(x - 4) \\[1em] \Rightarrow 9x - 36\end{aligned}$$

Hence, the simplified expression = $\bm{9x - 36}$.

Solving,

$$\begin{aligned}\Rightarrow 3x \times 5x^2y \times 2y\\[1em] \Rightarrow (3 \times 5 \times 2) \times (x \times x^2 \times y \times y)\\[1em] \Rightarrow 30x^3y^2\end{aligned}$$

Hence, $\bm{3x \times 5x^2y \times 2y = 30x^3y^2}$.

Solving,

$$\begin{aligned}\Rightarrow 5 \times 3a \times 2ab^2 \\[1em] \Rightarrow (5 \times 3 \times 2) \times (a \times a \times b^2) \\[1em] \Rightarrow 30a^2b^2\end{aligned}$$

Hence, $\bm{5 \times 3a \times 2ab^2 = 30a^2b^2}$.

Solving,

$$\begin{aligned}\Rightarrow (5x + 2y) \times 3xy \\[1em] \Rightarrow 5x \times 3xy + 2y \times 3xy \\[1em] \Rightarrow 15x^2y + 6xy^2\end{aligned}$$

Hence, $\bm{(5x + 2y) \times 3xy = 15x^2y + 6xy^2}$.

Solving,

$$\begin{aligned}\Rightarrow (6a - 5b) \times (-2a) \\[1em] \Rightarrow 6a \times (-2a) - 5b \times (-2a) \\[1em] \Rightarrow -12a^2 + 10ab\end{aligned}$$

Hence, $\bm{(6a - 5b) \times (-2a) = -12a^2 + 10ab}$.

Solving,

$$\begin{aligned}\Rightarrow (4a + 5b)(4a - 5b) \\[1em] \Rightarrow 4a(4a - 5b) + 5b(4a - 5b) \\[1em] \Rightarrow 16a^2 - 20ab + 20ab - 25b^2 \\[1em] \Rightarrow 16a^2 - 25b^2\end{aligned}$$

Hence, $\bm{(4a + 5b)(4a - 5b) = 16a^2 - 25b^2}$.

Solving,

$$\begin{aligned}\Rightarrow (9xy + 2y^2)(2x - 3y) \\[1em] \Rightarrow 9xy(2x - 3y) + 2y^2(2x - 3y) \\[1em] \Rightarrow 18x^2y - 27xy^2 + 4xy^2 - 6y^3 \\[1em] \Rightarrow 18x^2y - 23xy^2 - 6y^3\end{aligned}$$

Hence, $\bm{(9xy + 2y^2)(2x - 3y) = 18x^2y - 23xy^2 - 6y^3}$.

Solving,

$$\begin{aligned}\Rightarrow (-3m^2n + 5mn - 4mn^2) \times 6m^2n \\[1em] \Rightarrow -3m^2n \times 6m^2n + 5mn \times 6m^2n - 4mn^2 \times 6m^2n \\[1em] \Rightarrow -18m^4n^2 + 30m^3n^2 - 24m^3n^3\end{aligned}$$

Hence, $\bm{(-3m^2n + 5mn - 4mn^2) \times 6m^2n = -18m^4n^2 + 30m^3n^2 - 24m^3n^3}$.

Solving,

$$\begin{aligned}\Rightarrow (6xy^2 - 7x^2y^2 + 10x^3) \times (-3x^2y^3) \\[1em] \Rightarrow 6xy^2 \times (-3x^2y^3) - 7x^2y^2 \times (-3x^2y^3) + 10x^3 \times (-3x^2y^3) \\[1em] \Rightarrow -18x^3y^5 + 21x^4y^5 - 30x^5y^3\end{aligned}$$

Hence, $\bm{(6xy^2 - 7x^2y^2 + 10x^3) \times (-3x^2y^3) = -18x^3y^5 + 21x^4y^5 - 30x^5y^3}$.

$$\begin{array}{rrrr} & 3a & + & 2b \\ \times & & - & 3xy \\ \hline - & 9axy & - & 6bxy \\ \hline \end{array}$$

Hence, $\bm{(3a + 2b)(-3xy) = -9axy - 6bxy}$.

$$\begin{array}{rrrr} & 9x & - & 5y \\ \times & & - & 3xy \\ \hline - & 27x^2y & + & 15xy^2 \\ \hline \end{array}$$

Hence, $\bm{(9x - 5y)(-3xy) = -27x^2y + 15xy^2}$.

$$\begin{array}{rrrrrr} & 3xy & - & 2x^2 & - & 6x \\ \times & & & & - & 5x^2y \\ \hline - & 15x^3y^2 & + & 10x^4y & + & 30x^3y \\ \hline \end{array}$$

Hence, $\bm{(3xy - 2x^2 - 6x)(-5x^2y) = -15x^3y^2 + 10x^4y + 30x^3y}$.

$$\begin{array}{rrrrrrrr} & & & a & + & b & & \\ \times & & & a & + & b & & \\ \hline & & & a^2 & + & ab & & \\ & & + & & & ab & + & b^2 \\ \hline & & & a^2 & + & 2ab & + & b^2 \\ \hline \end{array}$$

Hence, $\bm{(a + b)(a + b) = a^2 + 2ab + b^2}$.

$$\begin{array}{rrrrrrrr} & & & & & ax & - & b \\ \times & & & & & 2ax & + & 2b^2 \\ \hline & & & & & 2a^2x^2 & - & 2abx \\ & & & & + & 2ab^2x & - & 2b^3 \\ \hline & 2a^2x^2 & - & 2abx & + & 2ab^2x & - & 2b^3 \\ \hline \end{array}$$

Hence, $\bm{(ax - b)(2ax + 2b^2) = 2a^2x^2 - 2abx + 2ab^2x - 2b^3}$.

$$\begin{array}{rrrrrrrrrr} & & & & & 2a & - & b & + & 3c \\ \times & & & & & & & 2a & - & 4b \\ \hline & & & & & 4a^2 & - & 2ab & + & 6ac \\ & & & & - & 8ab & + & 4b^2 & - & 12bc \\ \hline & 4a^2 & - & 10ab & + & 6ac & + & 4b^2 & - & 12bc \\ \hline \end{array}$$

Hence, $\bm{(2a - b + 3c)(2a - 4b) = 4a^2 - 10ab + 6ac + 4b^2 - 12bc}$.

$$\begin{array}{rrrrrrrrrr} & & & & & 3m^2 & + & 6m & - & 2n \\ \times & & & & & & & 5n & - & 3m \\ \hline & & & & & 15m^2n & + & 30mn & - & 10n^2 \\ & & & & - & 9m^3 & - & 18m^2 & + & 6mn \\ \hline - & 9m^3 & - & 18m^2 & + & 15m^2n & + & 36mn & - & 10n^2 \\ \hline \end{array}$$

Hence, $\bm{(3m^2 + 6m - 2n)(5n - 3m) = 15m^2n + 36mn - 10n^2 - 9m^3 - 18m^2}$.

$$\begin{array}{rrrrrrrrrr} & 6 & - & 3x & + & 2x^2 & & & & \\ \times & 1 & + & 5x & - & x^2 & & & & \\ \hline & 6 & - & 3x & + & 2x^2 & & & & \\ + & & & 30x & - & 15x^2 & + & 10x^3 & & \\ - & & & & & 6x^2 & + & 3x^3 & - & 2x^4 \\ \hline & 6 & + & 27x & - & 19x^2 & + & 13x^3 & - & 2x^4 \\ \hline \end{array}$$

Hence, $\bm{(6 - 3x + 2x^2)(1 + 5x - x^2) = 6 + 27x - 19x^2 + 13x^3 - 2x^4}$.

$$\begin{array}{rrrrrrrrrrrr} & & & & & 4x^3 & - & 10x^2 & + & 6x & - & 8 \\ \times & & & & & & & -x^2 & + & 2x & + & 3 \\ \hline & & & & & 12x^3 & - & 30x^2 & + & 18x & - & 24 \\ + & & & 8x^4 & - & 20x^3 & + & 12x^2 & - & 16x & & \\ - & 4x^5 & + & 10x^4 & - & 6x^3 & + & 8x^2 & & & & \\ \hline - & 4x^5 & + & 18x^4 & - & 14x^3 & - & 10x^2 & + & 2x & - & 24 \\ \hline \end{array}$$

Hence,

$\bm{(4x^3 - 10x^2 + 6x - 8)(3 + 2x - x^2) = -4x^5 + 18x^4 - 14x^3 - 10x^2 + 2x - 24}.$

Solving,

$$\begin{aligned}\Rightarrow (c + 5)(c - 3) \\[1em] \Rightarrow c(c - 3) + 5(c - 3) \\[1em] \Rightarrow c^2 - 3c + 5c - 15 \\[1em] \Rightarrow c^2 + 2c - 15\end{aligned}$$

Hence, $\bm{(c + 5)(c - 3) = c^2 + 2c - 15}$.

Solving,

$$\begin{aligned}\Rightarrow (3c - 5d)(4c - 6d) \\[1em] \Rightarrow 3c(4c - 6d) - 5d(4c - 6d) \\[1em] \Rightarrow 12c^2 - 18cd - 20cd + 30d^2 \\[1em] \Rightarrow 12c^2 - 38cd + 30d^2\end{aligned}$$

Hence, $\bm{(3c - 5d)(4c - 6d) = 12c^2 - 38cd + 30d^2}$.

Solving,

$$\begin{aligned}\Rightarrow \left(\frac{1}{2}a + \frac{1}{2}b\right)\left(\frac{1}{2}a - \frac{1}{2}b\right) \\[1em] \Rightarrow \frac{1}{2}a\left(\frac{1}{2}a - \frac{1}{2}b\right) + \frac{1}{2}b\left(\frac{1}{2}a - \frac{1}{2}b\right) \\[1em] \Rightarrow \frac{1}{4}a^2 - \frac{1}{4}ab + \frac{1}{4}ab - \frac{1}{4}b^2 \\[1em] \Rightarrow \frac{1}{4}a^2 - \frac{1}{4}b^2\end{aligned}$$

Hence, $\bm{\left(\frac{1}{2}a + \frac{1}{2}b\right)\left(\frac{1}{2}a - \frac{1}{2}b\right) = \frac{1}{4}a^2 - \frac{1}{4}b^2}$.

Solving,

$$\begin{aligned}\Rightarrow (a^2 + 2ab + b^2)(a + b) \\[1em] \Rightarrow a(a^2 + 2ab + b^2) + b(a^2 + 2ab + b^2) \\[1em] \Rightarrow a^3 + 2a^2b + ab^2 + a^2b + 2ab^2 + b^3 \\[1em] \Rightarrow a^3 + 3a^2b + 3ab^2 + b^3\end{aligned}$$

Hence, $\bm{(a^2 + 2ab + b^2)(a + b) = a^3 + 3a^2b + 3ab^2 + b^3}$.

Solving,

$$\begin{aligned}\Rightarrow (3x - 1)(4x^3 - 2x^2 + 6x - 3) \\[1em] \Rightarrow 3x(4x^3 - 2x^2 + 6x - 3) - 1(4x^3 - 2x^2 + 6x - 3) \\[1em] \Rightarrow 12x^4 - 6x^3 + 18x^2 - 9x - 4x^3 + 2x^2 - 6x + 3 \\[1em] \Rightarrow 12x^4 - 10x^3 + 20x^2 - 15x + 3\end{aligned}$$

Hence, $\bm{(3x - 1)(4x^3 - 2x^2 + 6x - 3) = 12x^4 - 10x^3 + 20x^2 - 15x + 3}$.

Solving,

$$\begin{aligned}\Rightarrow (4m - 2)(m^2 + 5m - 6) \\[1em] \Rightarrow 4m(m^2 + 5m - 6) - 2(m^2 + 5m - 6) \\[1em] \Rightarrow 4m^3 + 20m^2 - 24m - 2m^2 - 10m + 12 \\[1em] \Rightarrow 4m^3 + 18m^2 - 34m + 12\end{aligned}$$

Hence, $\bm{(4m - 2)(m^2 + 5m - 6) = 4m^3 + 18m^2 - 34m + 12}$.

Solving,

$$\begin{aligned}\Rightarrow (8 - 12x + 7x^2 - 6x^3)(5 - 2x) \\[1em] \Rightarrow 5(8 - 12x + 7x^2 - 6x^3) - 2x(8 - 12x + 7x^2 - 6x^3) \\[1em] \Rightarrow 40 - 60x + 35x^2 - 30x^3 - 16x + 24x^2 - 14x^3 + 12x^4 \\[1em] \Rightarrow 40 - 76x + 59x^2 - 44x^3 + 12x^4\end{aligned}$$

Hence, $\bm{(8 - 12x + 7x^2 - 6x^3)(5 - 2x) = 40 - 76x + 59x^2 - 44x^3 + 12x^4}$.

Solving,

$$\begin{aligned}\Rightarrow (4x^2 - 4x + 1)(2x^3 - 3x^2 + 2) \\[1em] \Rightarrow 4x^2(2x^3 - 3x^2 + 2) - 4x(2x^3 - 3x^2 + 2) + 1(2x^3 - 3x^2 + 2) \\[1em] \Rightarrow 8x^5 - 12x^4 + 8x^2 - 8x^4 + 12x^3 - 8x + 2x^3 - 3x^2 + 2 \\[1em] \Rightarrow 8x^5 + (-12 - 8)x^4 + (12 + 2)x^3 + (8 - 3)x^2 - 8x + 2 \\[1em] \Rightarrow 8x^5 - 20x^4 + 14x^3 + 5x^2 - 8x + 2\end{aligned}$$

Hence, $\bm{(4x^2 - 4x + 1)(2x^3 - 3x^2 + 2) = 8x^5 - 20x^4 + 14x^3 + 5x^2 - 8x + 2}$.

Solving,

$$\begin{aligned}\Rightarrow (6p^2 - 8pq + 2q^2) \times (-5p) \\[1em] \Rightarrow 6p^2 \times (-5p) - 8pq \times (-5p) + 2q^2 \times (-5p) \\[1em] \Rightarrow -30p^3 + 40p^2q - 10pq^2\end{aligned}$$

Hence, $\bm{(6p^2 - 8pq + 2q^2) \times (-5p) = -30p^3 + 40p^2q - 10pq^2}$.

Solving,

$$\begin{aligned}\Rightarrow -4y(15x + 12y - 8z)(x - 2y) \\[1em] \Rightarrow (-60xy - 48y^2 + 32yz)(x - 2y) \\[1em] \Rightarrow -60xy(x - 2y) - 48y^2(x - 2y) + 32yz(x - 2y) \\[1em] \Rightarrow -60x^2y + 120xy^2 - 48xy^2 + 96y^3 + 32xyz - 64y^2z \\[1em] \Rightarrow -60x^2y + 72xy^2 + 32xyz + 96y^3 - 64y^2z\end{aligned}$$

Hence, $\bm{-4y(15x + 12y - 8z)(x - 2y) = -60x^2y + 72xy^2 + 32xyz + 96y^3 - 64y^2z}$.

Solving,

$$\begin{aligned}\Rightarrow (a^2 + b^2 + c^2 - ab - bc - ca)(a + b + c) \\[1em] \Rightarrow a(a^2 + b^2 + c^2 - ab - bc - ca) + b(a^2 + b^2 + c^2 - ab - bc - ca) + c(a^2 + b^2 + c^2 - ab - bc - ca) \\[1em] \Rightarrow a^3 + ab^2 + ac^2 - a^2b - abc - a^2c + a^2b + b^3 + bc^2 - ab^2 - b^2c - abc + a^2c + b^2c + c^3 - abc - bc^2 - ac^2 \\[1em] \Rightarrow a^3 + b^3 + c^3 - 3abc\end{aligned}$$

Hence, $\bm{(a^2 + b^2 + c^2 - ab - bc - ca)(a + b + c) = a^3 + b^3 + c^3 - 3abc}$.

(i) $(a + b)(a - b)$

$$\begin{aligned}\Rightarrow a(a - b) + b(a - b) \\[1em] \Rightarrow a^2 - ab + ab - b^2 \\[1em] \Rightarrow a^2 - b^2\end{aligned}$$

Hence, $\bm{(a + b)(a - b) = a^2 - b^2}$.

(ii) $(a^2 + b^2)(a + b)(a - b)$

$$\begin{aligned}\Rightarrow (a^2 + b^2)(a^2 - b^2) \quad \text{[Using the result of (i)]} \\[1em] \Rightarrow a^2(a^2 - b^2) + b^2(a^2 - b^2) \\[1em] \Rightarrow a^4 - a^2b^2 + a^2b^2 - b^4 \\[1em] \Rightarrow a^4 - b^4\end{aligned}$$

Hence, $\bm{(a^2 + b^2)(a + b)(a - b) = a^4 - b^4}$.

(iii) $(a^4 + b^4)(a^2 + b^2)(a + b)(a - b)$

$$\begin{aligned}\Rightarrow (a^4 + b^4)(a^4 - b^4) \quad \text{[Using the result of (ii)]} \\[1em] \Rightarrow a^4(a^4 - b^4) + b^4(a^4 - b^4) \\[1em] \Rightarrow a^8 - a^4b^4 + a^4b^4 - b^8 \\[1em] \Rightarrow a^8 - b^8\end{aligned}$$

Hence, $\bm{(a^4 + b^4)(a^2 + b^2)(a + b)(a - b) = a^8 - b^8}$.

Solving,

$$\begin{aligned}\Rightarrow (3x - 2y)(4x + 3y) \\[1em] \Rightarrow 3x(4x + 3y) - 2y(4x + 3y) \\[1em] \Rightarrow 12x^2 + 9xy - 8xy - 6y^2 \\[1em] \Rightarrow 12x^2 + xy - 6y^2\end{aligned}$$

Hence, $\bm{(3x - 2y)(4x + 3y) = 12x^2 + xy - 6y^2}$.

Solving,

$$\begin{aligned}\Rightarrow (3x - 2y)(4x + 3y)(8x - 5y) \\[1em] \Rightarrow (12x^2 + xy - 6y^2)(8x - 5y) \quad \text{[Using the result of (i)]} \\[1em] \Rightarrow 8x(12x^2 + xy - 6y^2) - 5y(12x^2 + xy - 6y^2) \\[1em] \Rightarrow 96x^3 + 8x^2y - 48xy^2 - 60x^2y - 5xy^2 + 30y^3 \\[1em] \Rightarrow 96x^3 - 52x^2y - 53xy^2 + 30y^3\end{aligned}$$

Hence, $\bm{(3x - 2y)(4x + 3y)(8x - 5y) = 96x^3 - 52x^2y - 53xy^2 + 30y^3}$.

Solving,

$$\begin{aligned}\Rightarrow (a + 5)(3a - 2)(5a + 1) \\[1em] \Rightarrow [a(3a - 2) + 5(3a - 2)](5a + 1) \\[1em] \Rightarrow (3a^2 - 2a + 15a - 10)(5a + 1) \\[1em] \Rightarrow (3a^2 + 13a - 10)(5a + 1) \\[1em] \Rightarrow 5a(3a^2 + 13a - 10) + 1(3a^2 + 13a - 10) \\[1em] \Rightarrow 15a^3 + 65a^2 - 50a + 3a^2 + 13a - 10 \\[1em] \Rightarrow 15a^3 + 68a^2 - 37a - 10\end{aligned}$$

Hence, $\bm{(a + 5)(3a - 2)(5a + 1) = 15a^3 + 68a^2 - 37a - 10}$.

Solving $(a + 1)(a^2 - a + 1)$, we get :

$$\begin{aligned}\Rightarrow (a + 1)(a^2 - a + 1) \\[1em] \Rightarrow a(a^2 - a + 1) + 1(a^2 - a + 1) \\[1em] \Rightarrow a^3 - a^2 + a + a^2 - a + 1 \\[1em] \Rightarrow a^3 + 1\end{aligned}$$

Solving $(a - 1)(a^2 + a + 1)$, we get :

$$\begin{aligned}\Rightarrow (a - 1)(a^2 + a + 1) \\[1em] \Rightarrow a(a^2 + a + 1) - 1(a^2 + a + 1) \\[1em] \Rightarrow a^3 + a^2 + a - a^2 - a - 1 \\[1em] \Rightarrow a^3 - 1\end{aligned}$$

Solving, $(a + 1)(a^2 - a + 1) + (a - 1)(a^2 + a + 1)$

$$\begin{aligned}\Rightarrow (a + 1)(a^2 - a + 1) + (a - 1)(a^2 + a + 1) \\[1em] \Rightarrow (a^3 + 1) + (a^3 - 1) \\[1em] \Rightarrow 2a^3\end{aligned}$$

Hence, $\bm{(a + 1)(a^2 - a + 1) + (a - 1)(a^2 + a + 1) = 2a^3}$.

Solving,

$$\begin{aligned}\Rightarrow (5m - 2n)(5m + 2n)(25m^2 + 4n^2) \\[1em] \Rightarrow [5m(5m + 2n) - 2n(5m + 2n)](25m^2 + 4n^2) \\[1em] \Rightarrow (25m^2 + 10mn - 10mn - 4n^2)(25m^2 + 4n^2) \\[1em] \Rightarrow (25m^2 - 4n^2)(25m^2 + 4n^2) \\[1em] \Rightarrow 25m^2(25m^2 + 4n^2) - 4n^2(25m^2 + 4n^2) \\[1em] \Rightarrow 625m^4 + 100m^2n^2 - 100m^2n^2 - 16n^4 \\[1em] \Rightarrow 625m^4 - 16n^4\end{aligned}$$

Hence, $\bm{(5m - 2n)(5m + 2n)(25m^2 + 4n^2) = 625m^4 - 16n^4}$.

Solving,

$$\begin{aligned}\Rightarrow mn^4 \times m^3n \times 5m^2n^3 \\[1em] \Rightarrow 5 \times (m \times m^3 \times m^2) \times (n^4 \times n \times n^3) \\[1em] \Rightarrow 5m^6n^8\end{aligned}$$

Hence, $\bm{mn^4 \times m^3n \times 5m^2n^3 = 5m^6n^8}$.

Solving,

$$\begin{aligned}\Rightarrow 2mnpq \times 4mnpq \times 5mnpq \\[1em] \Rightarrow (2 \times 4 \times 5) \times m^3n^3p^3q^3 \\[1em] \Rightarrow 40m^3n^3p^3q^3\end{aligned}$$

Hence, $\bm{2mnpq \times 4mnpq \times 5mnpq = 40m^3n^3p^3q^3}$.

Solving,

$$\begin{aligned}\Rightarrow (pq - pm) \times p^2m \\[1em] \Rightarrow pq \times p^2m - pm \times p^2m \\[1em] \Rightarrow p^3qm - p^3m^2\end{aligned}$$

Hence, $\bm{(pq - pm) \times p^2m = p^3qm - p^3m^2}$.

Solving,

$$\begin{aligned}\Rightarrow (x^3 - 3y^3) \times 4x^2y^2 \\[1em] \Rightarrow x^3 \times 4x^2y^2 - 3y^3 \times 4x^2y^2 \\[1em] \Rightarrow 4x^5y^2 - 12x^2y^5\end{aligned}$$

Hence, $\bm{(x^3 - 3y^3) \times 4x^2y^2 = 4x^5y^2 - 12x^2y^5}$.

Solving,

$$\begin{aligned}\Rightarrow (a^3 - 4ab) \times 2a^2b \\[1em] \Rightarrow a^3 \times 2a^2b - 4ab \times 2a^2b \\[1em] \Rightarrow 2a^5b - 8a^3b^2\end{aligned}$$

Hence, $\bm{(a^3 - 4ab) \times 2a^2b = 2a^5b - 8a^3b^2}$.

Solving,

$$\begin{aligned}\Rightarrow (x^2 + 5yx - 3y^2) \times 2x^2y \\[1em] \Rightarrow x^2 \times 2x^2y + 5xy \times 2x^2y - 3y^2 \times 2x^2y \\[1em] \Rightarrow 2x^4y + 10x^3y^2 - 6x^2y^3\end{aligned}$$

Hence, $\bm{(x^2 + 5yx - 3y^2) \times 2x^2y = 2x^4y + 10x^3y^2 - 6x^2y^3}$.

Solving,

$$\begin{aligned}\Rightarrow (2x + 3y)(2x + 3y) \\[1em] \Rightarrow 2x(2x + 3y) + 3y(2x + 3y) \\[1em] \Rightarrow 4x^2 + 6xy + 6xy + 9y^2 \\[1em] \Rightarrow 4x^2 + 12xy + 9y^2\end{aligned}$$

Hence, $\bm{(2x + 3y)(2x + 3y) = 4x^2 + 12xy + 9y^2}$.

Solving,

$$\begin{aligned}\Rightarrow (2x - 3y)(2x + 3y) \\[1em] \Rightarrow 2x(2x + 3y) - 3y(2x + 3y) \\[1em] \Rightarrow 4x^2 + 6xy - 6xy - 9y^2 \\[1em] \Rightarrow 4x^2 - 9y^2\end{aligned}$$

Hence, $\bm{(2x - 3y)(2x + 3y) = 4x^2 - 9y^2}$.

Solving,

$$\begin{aligned}\Rightarrow (2x + 3y)(2x - 3y) \\[1em] \Rightarrow 2x(2x - 3y) + 3y(2x - 3y) \\[1em] \Rightarrow 4x^2 - 6xy + 6xy - 9y^2 \\[1em] \Rightarrow 4x^2 - 9y^2\end{aligned}$$

Hence, $\bm{(2x + 3y)(2x - 3y) = 4x^2 - 9y^2}$.

Solving,

$$\begin{aligned}\Rightarrow (2x - 3y)(2x - 3y) \\[1em] \Rightarrow 2x(2x - 3y) - 3y(2x - 3y) \\[1em] \Rightarrow 4x^2 - 6xy - 6xy + 9y^2 \\[1em] \Rightarrow 4x^2 - 12xy + 9y^2\end{aligned}$$

Hence, $\bm{(2x - 3y)(2x - 3y) = 4x^2 - 12xy + 9y^2}$.

Solving,

$$\begin{aligned}\Rightarrow (-2x + 3y)(2x - 3y) \\[1em] \Rightarrow -2x(2x - 3y) + 3y(2x - 3y) \\[1em] \Rightarrow -4x^2 + 6xy + 6xy - 9y^2 \\[1em] \Rightarrow -4x^2 + 12xy - 9y^2\end{aligned}$$

Hence, $\bm{(-2x + 3y)(2x - 3y) = -4x^2 + 12xy - 9y^2}$.

Solving,

$$\begin{aligned}\Rightarrow (xy + 2b)(xy - 2b) \\[1em] \Rightarrow xy(xy - 2b) + 2b(xy - 2b) \\[1em] \Rightarrow x^2y^2 - 2bxy + 2bxy - 4b^2 \\[1em] \Rightarrow x^2y^2 - 4b^2\end{aligned}$$

Hence, $\bm{(xy + 2b)(xy - 2b) = x^2y^2 - 4b^2}$.

Solving,

$$\begin{aligned}\Rightarrow (x - a)(x + 3b) \\[1em] \Rightarrow x(x + 3b) - a(x + 3b) \\[1em] \Rightarrow x^2 + 3bx - ax - 3ab\end{aligned}$$

Hence, $\bm{(x - a)(x + 3b) = x^2 + 3bx - ax - 3ab}$.

Solving,

$$\begin{aligned}\Rightarrow (2x + 5y + 6)(3x + y - 8) \\[1em] \Rightarrow 2x(3x + y - 8) + 5y(3x + y - 8) + 6(3x + y - 8) \\[1em] \Rightarrow 6x^2 + 2xy - 16x + 15xy + 5y^2 - 40y + 18x + 6y - 48 \\[1em] \Rightarrow 6x^2 + (2 + 15)xy + 5y^2 + (-16 + 18)x + (-40 + 6)y - 48 \\[1em] \Rightarrow 6x^2 + 17xy + 5y^2 + 2x - 34y - 48\end{aligned}$$

Hence, $\bm{(2x + 5y + 6)(3x + y - 8) = 6x^2 + 17xy + 5y^2 + 2x - 34y - 48}$.

Solving,

$$\begin{aligned}\Rightarrow (3x - 5y + 2)(5x - 4y - 3) \\[1em] \Rightarrow 3x(5x - 4y - 3) - 5y(5x - 4y - 3) + 2(5x - 4y - 3) \\[1em] \Rightarrow 15x^2 - 12xy - 9x - 25xy + 20y^2 + 15y + 10x - 8y - 6 \\[1em] \Rightarrow 15x^2 + (-12 - 25)xy + 20y^2 + (-9 + 10)x + (15 - 8)y - 6 \\[1em] \Rightarrow 15x^2 - 37xy + 20y^2 + x + 7y - 6\end{aligned}$$

Hence, $\bm{(3x - 5y + 2)(5x - 4y - 3) = 15x^2 - 37xy + 20y^2 + x + 7y - 6}$.

Solving,

$$\begin{aligned}\Rightarrow (6x - 2y)(3x - y) \\[1em] \Rightarrow 6x(3x - y) - 2y(3x - y) \\[1em] \Rightarrow 18x^2 - 6xy - 6xy + 2y^2 \\[1em] \Rightarrow 18x^2 - 12xy + 2y^2\end{aligned}$$

Hence, $\bm{(6x - 2y)(3x - y) = 18x^2 - 12xy + 2y^2}$.

Solving,

$$\begin{aligned}\Rightarrow (1 + 6x^2 - 4x^3)(-1 + 3x - 3x^2) \\[1em] \Rightarrow 1(-1 + 3x - 3x^2) + 6x^2(-1 + 3x - 3x^2) - 4x^3(-1 + 3x - 3x^2) \\[1em] \Rightarrow -1 + 3x - 3x^2 - 6x^2 + 18x^3 - 18x^4 + 4x^3 - 12x^4 + 12x^5 \\[1em] \Rightarrow -1 + 3x + (-3 - 6)x^2 + (18 + 4)x^3 + (-18 - 12)x^4 + 12x^5 \\[1em] \Rightarrow -1 + 3x - 9x^2 + 22x^3 - 30x^4 + 12x^5\end{aligned}$$

Hence, $\bm{(1 + 6x^2 - 4x^3)(-1 + 3x - 3x^2) = -1 + 3x - 9x^2 + 22x^3 - 30x^4 + 12x^5}$.

Solving,

$\Rightarrow \frac{-16ab^2c}{6abc} = \frac{-8b}{3}$

Hence, $\bm{\frac{-16ab^2c}{6abc} = \frac{-8b}{3}}$.

Solving,

$\Rightarrow \frac{25x^2y}{-5y^2} = \frac{-5x^2}{y}$

Hence, $\bm{\frac{25x^2y}{-5y^2} = \frac{-5x^2}{y}}$.

Solving,

$$\begin{aligned}\Rightarrow \frac{8x + 24}{4}\\[1em] \Rightarrow \frac{8x}{4} + \frac{24}{4}\\[1em] \Rightarrow 2x + 6\end{aligned}$$

Hence, $\bm{\frac{8x + 24}{4} = 2x + 6}$.

Solving,

$$\begin{aligned}\Rightarrow \frac{4a^2 - a}{-a}\\[1em] \Rightarrow \frac{4a^2}{-a} - \frac{a}{-a}\\[1em] \Rightarrow -4a - (-1)\\[1em] \Rightarrow -4a + 1\end{aligned}$$

Hence, $\bm{\frac{4a^2 - a}{-a} = -4a + 1}$.

Solving,

$$\begin{aligned}\Rightarrow \frac{8m - 16}{-8}\\[1em] \Rightarrow \frac{8m}{-8} - \frac{16}{-8}\\[1em] \Rightarrow -m - (-2)\\[1em] \Rightarrow -m + 2\end{aligned}$$

Hence, $\bm{\frac{8m - 16}{-8} = -m + 2}$.

Solving,

$$\begin{aligned}\Rightarrow \frac{-50 + 40p}{10p}\\[1em] \Rightarrow \frac{-50}{10p} + \frac{40p}{10p}\\[1em] \Rightarrow -\frac{5}{p} + 4\end{aligned}$$

Hence, $\bm{\frac{-50 + 40p}{10p} = -\frac{5}{p} + 4}$.

Solving,

$$\begin{aligned}\Rightarrow \frac{4x^3 - 2x^2}{-x}\\[1em] \Rightarrow \frac{4x^3}{-x} - \frac{2x^2}{-x}\\[1em] \Rightarrow -4x^2 - (-2x)\\[1em] \Rightarrow -4x^2 + 2x\end{aligned}$$

Hence, $\bm{\frac{4x^3 - 2x^2}{-x} = -4x^2 + 2x}$.

Solving,

$$\begin{aligned}\Rightarrow \frac{10a^3 - 15a^2b}{-5a^2}\\[1em] \Rightarrow \frac{10a^3}{-5a^2} - \frac{15a^2b}{-5a^2}\\[1em] \Rightarrow -2a - (-3b)\\[1em] \Rightarrow -2a + 3b\end{aligned}$$

Hence, $\bm{\frac{10a^3 - 15a^2b}{-5a^2} = -2a + 3b}$.

Solving,

$$\begin{aligned}\Rightarrow \frac{12x^3y - 8x^2y^2 + 4x^2y^3}{4xy}\\[1em] \Rightarrow \frac{12x^3y}{4xy} - \frac{8x^2y^2}{4xy} + \frac{4x^2y^3}{4xy}\\[1em] \Rightarrow 3x^2 - 2xy + xy^2\end{aligned}$$

Hence, $\bm{\frac{12x^3y - 8x^2y^2 + 4x^2y^3}{4xy} = 3x^2 - 2xy + xy^2}$.

Solving,

$$\begin{aligned}\Rightarrow \frac{9a^4b - 15a^3b^2 + 12a^2b^3}{-3a^2b}\\[1em] \Rightarrow \frac{9a^4b}{-3a^2b} - \frac{15a^3b^2}{-3a^2b} + \frac{12a^2b^3}{-3a^2b}\\[1em] \Rightarrow -3a^2 - (-5ab) + (-4b^2)\\[1em] \Rightarrow -3a^2 + 5ab - 4b^2\end{aligned}$$

Hence, $\bm{\frac{9a^4b - 15a^3b^2 + 12a^2b^3}{-3a^2b} = -3a^2 + 5ab - 4b^2}$.

Solving,

$$\begin{array}{l} n - 1\ \overline{\smash{\big)}\ n^2 - 2n + 1\ \smash{\big(}}\ n - 1 \\ \phantom{n - 1)}\underline{\underset{-}{}n^2 \underset{+}{-}\phantom{2}n\phantom{{} + 1}} \\ \phantom{n - 1\ \big)\ }\phantom{n^2\ {}}-n + 1 \\ \phantom{n --1))}\phantom{n^2\ {}}\underline{\underset{+}{-}n \underset{-}{+}1} \\ \phantom{n - 1}\phantom{++++}0 \end{array}$$

Hence, $\bm{\frac{n^2 - 2n + 1}{n - 1} = n - 1}$.

Solving,

$$\begin{array}{l} m - n\ \overline{\smash{\big)}\ m^2 - 2mn + n^2\ \smash{\big(}}\ m - n \\ \phantom{m - n)}\underline{\underset{-}{}m^2 \underset{+}{-}\phantom{2}mn\phantom{{} + n^2}} \\ \phantom{m - n\ \big)\ }\phantom{m^2\ {}}-mn + n^2 \\ \phantom{m - n)}\phantom{++(}\underline{\underset{+}{-}mn \underset{-}{+}n^2} \\ \phantom{m - n}\phantom{m^2 - 2mn + {}}0 \end{array}$$

Hence, $\bm{\frac{m^2 - 2mn + n^2}{m - n} = m - n}$.

Solving,

$$\begin{array}{l} 2a + 1\ \overline{\smash{\big)}\ 4a^2 + 4a + 1\ \smash{\big(}}\ 2a + 1 \\ \phantom{2a + 1)}\underline{\underset{-}{}4a^2 \underset{-}{+}2a\phantom{{} + 1}} \\ \phantom{2a + 1\ \big)\ }\phantom{4a^2 + {}}2a + 1 \\ \phantom{2a + 1)}\phantom{4a^2 + {}}\underline{\underset{-}{}2a \underset{-}{+}1} \\ \phantom{2a + 1}\phantom{4a^2 + 2a + {}}0 \end{array}$$

Hence, $\bm{\frac{4a^2 + 4a + 1}{2a + 1} = 2a + 1}$.

Solving,

$$\begin{array}{l} p + 2\ \overline{\smash{\big)}\ p^2 + 4p + 4\ \smash{\big(}}\ p + 2 \\ \phantom{p + 2)}\underline{\underset{-}{}p^2 \underset{-}{+}2p\phantom{{} + 4}} \\ \phantom{p + 2\ \big)\ }\phantom{p^2 + {}}2p + 4 \\ \phantom{p + 2)}\phantom{p^2 + {}}\underline{\underset{-}{}2p \underset{-}{+}4} \\ \phantom{p + 2}\phantom{p^2 + 2p + {}}0 \end{array}$$

Hence, $\bm{\frac{p^2 + 4p + 4}{p + 2} = p + 2}$.

Solving,

$$\begin{array}{l} x + 2y\ \overline{\smash{\big)}\ x^2 + 4xy + 4y^2\ \smash{\big(}}\ x + 2y \\ \phantom{x + 2y)}\underline{\underset{-}{}x^2 \underset{-}{+}2xy\phantom{{} + 4y^2}} \\ \phantom{x + 2y\ \big)\ }\phantom{x^2 + {}}2xy + 4y^2 \\ \phantom{x + 2y)}\phantom{x^2 + {}}\underline{\underset{-}{}2xy \underset{-}{+}4y^2} \\ \phantom{x + 2y}\phantom{x^2 + 2xy + {}}0 \end{array}$$

Hence, $\bm{\frac{x^2 + 4xy + 4y^2}{x + 2y} = x + 2y}$.

Solving,

$$\begin{array}{l} a - 4\ \overline{\smash{\big)}\ 2a^2 - 11a + 12\ \smash{\big(}}\ 2a - 3 \\ \phantom{a - 4)}\underline{\underset{-}{}2a^2 \underset{+}{-}\phantom{1}8a\phantom{{} + 12}} \\ \phantom{a - 4\ \big)\ }\phantom{2a^2\ {}}-3a + 12 \\ \phantom{a - 4)}\phantom{++()}\underline{\underset{+}{-}3a \underset{-}{+}12} \\ \phantom{a - 4}\phantom{2a^2 - 11a + {}}0 \end{array}$$

Hence, $\bm{\frac{2a^2 - 11a + 12}{a - 4} = 2a - 3}$.

Solving,

$$\begin{array}{l} 2x + 3\ \overline{\smash{\big)}\ 6x^2 + 5x - 6\ \smash{\big(}}\ 3x - 2 \\ \phantom{2x + 3)}\underline{\underset{-}{}6x^2 \underset{-}{+}9x\phantom{{} - 6}} \\ \phantom{2x + 3\ \big)\ }\phantom{6x^2\ {}}-4x - 6 \\ \phantom{2x + 3)}\phantom{++()}\underline{\underset{+}{-}4x \underset{+}{-}6} \\ \phantom{2x + 3}\phantom{6x^2 + 5x - {}}0 \end{array}$$

Hence, $\bm{\frac{6x^2 + 5x - 6}{2x + 3} = 3x - 2}$.

Solving,

$$\begin{array}{l} 2a - 5\ \overline{\smash{\big)}\ 8a^2 + \phantom{2}4a - 60\ \smash{\big(}}\ 4a + 12 \\ \phantom{2a - 5)}\underline{\underset{-}{}8a^2 \underset{+}{-}20a\phantom{{} - 60}} \\ \phantom{2a - 5\ \big)\ }\phantom{8a^2 + {}}24a - 60 \\ \phantom{2a - 5)}\phantom{8a^2 + {}}\underline{\underset{-}{}24a \underset{+}{-}60} \\ \phantom{2a - 5}\phantom{8a^2 + 24a - {}}0 \end{array}$$

Hence, $\bm{\frac{8a^2 + 4a - 60}{2a - 5} = 4a + 12}$.

Solving,

$$\begin{array}{l} 3x - 4y\ \overline{\smash{\big)}\ 9x^2 - 24xy + 16y^2\ \smash{\big(}}\ 3x - 4y \\ \phantom{3x - 4y)}\underline{\underset{-}{}9x^2 \underset{+}{-}12xy\phantom{{} + 16y^2}} \\ \phantom{3x - 4y\ \big)\ }\phantom{9x^2\ {}}-12xy + 16y^2 \\ \phantom{3x - 4y)}\phantom{++()}\underline{\underset{+}{-}12xy \underset{-}{+}16y^2} \\ \phantom{3x - 4y}\phantom{9x^2 - 24xy + {}}0 \end{array}$$

Hence, $\bm{\frac{9x^2 - 24xy + 16y^2}{3x - 4y} = 3x - 4y}$.

Solving,

$$\begin{array}{l} 5x + 7y\ \overline{\smash{\big)}\ 15x^2 + 31xy + 14y^2\ \smash{\big(}}\ 3x + 2y \\ \phantom{5x + 7y)}\underline{\underset{-}{}15x^2 \underset{-}{+}21xy\phantom{{} + 14y^2}} \\ \phantom{5x + 7y\ \big)\ }\phantom{15x^2 + {}}10xy + 14y^2 \\ \phantom{5x + 7y)}\phantom{15x^2 + {}}\underline{\underset{-}{}10xy \underset{-}{+}14y^2} \\ \phantom{5x + 7y}\phantom{15x^2 + 10xy + {}}0 \end{array}$$

Hence, $\bm{\frac{15x^2 + 31xy + 14y^2}{5x + 7y} = 3x + 2y}$.

Solving,

$$\begin{array}{l} 5a - b\ \overline{\smash{\big)}\ 35a^3 + \phantom{1}3a^2b - 2ab^2\ \smash{\big(}}\ 7a^2 + 2ab \\ \phantom{5a - b)}\underline{\underset{-}{}35a^3 \underset{+}{-}\phantom{1}7a^2b\phantom{{} - 2ab^2}} \\ \phantom{5a - b\ \big)\ }\phantom{35a^3 + {}}10a^2b - 2ab^2 \\ \phantom{5a - b)}\phantom{35a^3 + {}}\underline{\underset{-}{}10a^2b \underset{+}{-}2ab^2} \\ \phantom{5a - b}\phantom{35a^3 + 10a^2b - {}}0 \end{array}$$

Hence, $\bm{\frac{35a^3 + 3a^2b - 2ab^2}{5a - b} = 7a^2 + 2ab}$.

Solving,

$$\begin{array}{l} 3x - 2\ \overline{\smash{\big)}\ 6x^3 + \phantom{1}5x^2 - 21x + 10\ \smash{\big(}}\ 2x^2 + 3x - 5 \\ \phantom{3x - 2)}\underline{\underset{-}{}6x^3 \underset{+}{-}\phantom{1}4x^2\phantom{{} - 21x + 10}} \\ \phantom{3x - 2\ \big)\ }\phantom{++())}9x^2 - 21x + 10 \\ \phantom{3x - 2)}\phantom{++())}\underline{\underset{-}{}9x^2 \underset{+}{-}\phantom{1}6x\phantom{{} + 10}} \\ \phantom{3x - 2\ \big)\ }\phantom{++++()}-15x + 10 \\ \phantom{3x - 2)}\phantom{+++++()}\underline{\underset{+}{-}15x \underset{-}{+}10} \\ \phantom{3x - 2}\phantom{++++++++()}0 \end{array}$$

Hence, $\bm{\frac{6x^3 + 5x^2 - 21x + 10}{3x - 2} = 2x^2 + 3x - 5}$.

Area of a rectangle = length × breadth

∴ Breadth = $\frac{\text{Area}}{\text{Length}} = \frac{6x^2 - 4xy - 10y^2}{2x + 2y}$

Dividing $6x^2 - 4xy - 10y^2$ by $2x + 2y$:

$$\begin{array}{l} 2x + 2y\ \overline{\smash{\big)}\ 6x^2 - \phantom{1}4xy - 10y^2\ \smash{\big(}}\ 3x - 5y \\ \phantom{2x + 2y)}\underline{\underset{-}{}6x^2 \underset{-}{+}\phantom{1}6xy\phantom{{} - 10y^2}} \\ \phantom{2x + 2y\ \big)\ }\phantom{6x^2\ {}}-10xy - 10y^2 \\ \phantom{2x + 2y)}\phantom{++()}\underline{\underset{+}{-}10xy \underset{+}{-}10y^2} \\ \phantom{2x + 2y}\phantom{6x^2 - 10xy - {}}0 \end{array}$$

Hence, the breadth of the rectangle is $\bm{(3x - 5y)}$ unit.

Breadth = $\frac{\text{Area}}{\text{Length}} = \frac{25x^2 + 20xy + 3y^2}{5x + 3y}$

Dividing $25x^2 + 20xy + 3y^2$ by $5x + 3y$:

$$\begin{array}{l} 5x + 3y\ \overline{\smash{\big)}\ 25x^2 + 20xy + 3y^2\ \smash{\big(}}\ 5x + y \\ \phantom{5x + 3y)}\underline{\underset{-}{}25x^2 \underset{-}{+}15xy\phantom{{} + 3y^2}} \\ \phantom{5x + 3y\ \big)\ }\phantom{25x^2 + {}}5xy + 3y^2 \\ \phantom{5x + 3y)}\phantom{25x^2 + {}}\underline{\underset{-}{}5xy \underset{-}{+}3y^2} \\ \phantom{5x + 3y}\phantom{25x^2 + 5xy + {}}0 \end{array}$$

∴ Breadth = $5x + y$

$$\begin{aligned}\text{ Perimeter } = 2 \times \text{ (length + breadth) }\\[1em] = 2 \times [(5x + 3y) + (5x + y)]\\[1em] = 2 \times (10x + 4y)\\[1em] = 20x + 8y\end{aligned}$$

Hence, the breadth = $\bm{(5x + y)}$ unit and the perimeter = $\bm{(20x + 8y)}$ unit.

Solving,

$$\begin{aligned}\Rightarrow \frac{2m^3n^5}{-mn} \\[1em] \Rightarrow -2m^2n^4\end{aligned}$$

Hence, final result = $\bm{-2m^2n^4}$.

Solving,

$$\begin{aligned}\Rightarrow \frac{5x^2 - 3x}{x}\\[1em] \Rightarrow \frac{5x^2}{x} - \frac{3x}{x}\\[1em] \Rightarrow 5x - 3\end{aligned}$$

Hence, final result = $\bm{5x - 3}$.

Solving,

$$\begin{aligned}\Rightarrow \frac{10x^3y - 9xy^2 - 4x^2y^2}{xy}\\[1em] \Rightarrow \frac{10x^3y}{xy} - \frac{9xy^2}{xy} - \frac{4x^2y^2}{xy}\\[1em] \Rightarrow 10x^2 - 9y - 4xy\end{aligned}$$

Hence, final result = $\bm{10x^2 - 9y - 4xy}$.

Solving,

$$\begin{aligned}\Rightarrow \frac{3y^3 - 9ay^2 - 6ab^2y}{-3y}\\[1em] \Rightarrow \frac{3y^3}{-3y} - \frac{9ay^2}{-3y} - \frac{6ab^2y}{-3y}\\[1em] \Rightarrow -y^2 - (-3ay) - (-2ab^2)\\[1em] \Rightarrow -y^2 + 3ay + 2ab^2\end{aligned}$$

Hence, final result = $\bm{-y^2 + 3ay + 2ab^2}$.

Solving,

$$\begin{aligned}\Rightarrow \frac{x^5 - 15x^4 - 10x^2}{-5x^2}\\[1em] \Rightarrow \frac{x^5}{-5x^2} - \frac{15x^4}{-5x^2} - \frac{10x^2}{-5x^2}\\[1em] \Rightarrow -\frac{1}{5}x^3 - (-3x^2) - (-2)\\[1em] \Rightarrow -\frac{1}{5}x^3 + 3x^2 + 2\end{aligned}$$

Hence, final result = $\bm{-\frac{1}{5}x^3 + 3x^2 + 2}$.

Solving,

$$\begin{array}{l} 3a - 2x\ \overline{\smash{\big)}\ 12a^2 + \phantom{9}ax - 6x^2\ \smash{\big(}}\ 4a + 3x \\ \phantom{3a - 2x)}\underline{\underset{-}{}12a^2 \underset{+}{-}8ax\phantom{{} - 6x^2}} \\ \phantom{3a - 2x\ \big)\ }\phantom{12a^2 + {}}9ax - 6x^2 \\ \phantom{3a - 2x)}\phantom{12a^2 + {}}\underline{\underset{-}{}9ax \underset{+}{-}6x^2} \\ \phantom{3a - 2x}\phantom{12a^2 + 9ax - {}}0 \end{array}$$

Hence, final result = $\bm{4a + 3x}$.

Solving,

$$\begin{array}{l} 2x - 5y\ \overline{\smash{\big)}\ 6x^2 - \phantom{1}xy - 35y^2\ \smash{\big(}}\ 3x + 7y \\ \phantom{2x - 5y)}\underline{\underset{-}{}6x^2 \underset{+}{-}15xy\phantom{{} - 35y^2}} \\ \phantom{2x - 5y\ \big)\ }\phantom{6x^2 + {}}14xy - 35y^2 \\ \phantom{2x - 5y)}\phantom{6x^2 + {}}\underline{\underset{-}{}14xy \underset{+}{-}35y^2} \\ \phantom{2x - 5y}\phantom{6x^2 + 14xy - {}}0 \end{array}$$

Hence, final result = $\bm{3x + 7y}$.

Solving,

$$\begin{array}{l} x^2 - 4x + 3\ \overline{\smash{\big)}\ x^3 - 6x^2 + 11x - 6\ \smash{\big(}}\ x - 2 \\ \phantom{x^2 - 4x + 3)}\underline{\underset{-}{}x^3 \underset{+}{-}4x^2 \underset{-}{+}\phantom{1}3x\phantom{{} - 6}} \\ \phantom{x^2 - 4x + 3\ \big)\ }\phantom{x^3\ {}}-2x^2 + \phantom{1}8x - 6 \\ \phantom{x^2 - 4x + 3)}\phantom{++}\underline{\underset{+}{-}2x^2 \underset{-}{+}\phantom{1}8x \underset{+}{-}6} \\ \phantom{x^2 - 4x + 3}\phantom{x^3 - 2x^2 + 8x - {}}0 \end{array}$$

Hence, final result = $\bm{x - 2}$.

Solving,

$$\begin{array}{l} m^2 - m - 2\ \overline{\smash{\big)}\ m^3 - 4m^2 + \phantom{3}m + 6\ \smash{\big(}}\ m - 3 \\ \phantom{m^2 - m - 2)}\underline{\underset{-}{}m^3 \underset{+}{-}\phantom{4}m^2 \underset{+}{-}2m\phantom{{} + 6}} \\ \phantom{m^2 - m - 2\ \big)\ }\phantom{m^3\ {}}-3m^2 + 3m + 6 \\ \phantom{m^2 - m - 2)}\phantom{++()}\underline{\underset{+}{-}3m^2 \underset{-}{+}3m \underset{-}{+}6} \\ \phantom{m^2 - m - 2}\phantom{m^3 - 3m^2 + 3m + {}}0 \end{array}$$

Hence, final result = $\bm{m - 3}$.

Simplifying,

$$\begin{aligned}\Rightarrow \frac{x}{2} + \frac{x}{4}\\[1em] \Rightarrow \frac{2x + x}{4}[\text{ L.C.M. of 2 and 4 = 4 }]\\[1em] \Rightarrow \frac{3x}{4}\end{aligned}$$

Hence, $\bm{\frac{x}{2} + \frac{x}{4} = \frac{3x}{4}}$.

Simplifying,

$$\begin{aligned}\Rightarrow \frac{a}{10} + \frac{2a}{5}\\[1em] \Rightarrow \frac{a + 4a}{10}[\text{ L.C.M. of 10 and 5 = 10 }]\\[1em] \Rightarrow \frac{5a}{10}\\[1em] \Rightarrow \frac{a}{2}\end{aligned}$$

Hence, $\bm{\frac{a}{10} + \frac{2a}{5} = \frac{a}{2}}$.

Simplifying,

$$\begin{aligned}\Rightarrow \frac{y}{4} + \frac{3y}{5}\\[1em] \Rightarrow \frac{5y + 12y}{20}[\text{ L.C.M. of 4 and 5 = 20 }]\\[1em] \Rightarrow \frac{17y}{20}\end{aligned}$$

Hence, $\bm{\frac{y}{4} + \frac{3y}{5} = \frac{17y}{20}}$.

Simplifying,

$$\begin{aligned}\Rightarrow \frac{x}{2} - \frac{x}{8}\\[1em] \Rightarrow \frac{4x - x}{8}[\text{ L.C.M. of 2 and 8 = 8 }]\\[1em] \Rightarrow \frac{3x}{8}\end{aligned}$$

Hence, $\bm{\frac{x}{2} - \frac{x}{8} = \frac{3x}{8}}$.

Simplifying,

$$\begin{aligned}\Rightarrow \frac{3y}{4} - \frac{y}{5}\\[1em] \Rightarrow \frac{15y - 4y}{20}[\text{ L.C.M. of 4 and 5 = 20 }]\\[1em] \Rightarrow \frac{11y}{20}\end{aligned}$$

Hence, $\bm{\frac{3y}{4} - \frac{y}{5} = \frac{11y}{20}}$.

Simplifying,

$$\begin{aligned}\Rightarrow \frac{2p}{3} - \frac{3p}{5}\\[1em] \Rightarrow \frac{10p - 9p}{15}[\text{ L.C.M. of 3 and 5 = 15 }]\\[1em] \Rightarrow \frac{p}{15}\end{aligned}$$

Hence, $\bm{\frac{2p}{3} - \frac{3p}{5} = \frac{p}{15}}$.

Simplifying,

$$\begin{aligned}\Rightarrow \frac{k}{2} + \frac{k}{3} + \frac{2k}{5}\\[1em] \Rightarrow \frac{15k + 10k + 12k}{30}[\text{ L.C.M. of 2, 3 and 5 = 30 }]\\[1em] \Rightarrow \frac{37k}{30}\end{aligned}$$

Hence, $\bm{\frac{k}{2} + \frac{k}{3} + \frac{2k}{5} = \frac{37k}{30}}$.

Simplifying,

$$\begin{aligned}\Rightarrow \frac{2x}{5} + \frac{3x}{4} - \frac{3x}{5}\\[1em] \Rightarrow \frac{8x + 15x - 12x}{20}[\text{ L.C.M. of 5, 4 and 5 = 20 }]\\[1em] \Rightarrow \frac{11x}{20}\end{aligned}$$

Hence, $\bm{\frac{2x}{5} + \frac{3x}{4} - \frac{3x}{5} = \frac{11x}{20}}$.

Simplifying,

$$\begin{aligned}\Rightarrow \frac{4a}{7} - \frac{2a}{3} + \frac{a}{7}\\[1em] \Rightarrow \frac{12a - 14a + 3a}{21}[\text{ L.C.M. of 7, 3 and 7 = 21 }]\\[1em] \Rightarrow \frac{a}{21}\end{aligned}$$

Hence, $\bm{\frac{4a}{7} - \frac{2a}{3} + \frac{a}{7} = \frac{a}{21}}$.

Simplifying,

$$\begin{aligned}\Rightarrow \frac{2b}{5} - \frac{7b}{15} + \frac{13b}{3}\\[1em] \Rightarrow \frac{6b - 7b + 65b}{15}[\text{ L.C.M. of 5, 15 and 3 = 15 }]\\[1em] \Rightarrow \frac{64b}{15}\end{aligned}$$

Hence, $\bm{\frac{2b}{5} - \frac{7b}{15} + \frac{13b}{3} = \frac{64b}{15}}$.

Simplifying,

$$\begin{aligned}\Rightarrow \frac{6k}{7} - \left(\frac{8k}{9} - \frac{k}{3}\right)\\[1em] \Rightarrow \frac{6k}{7} - \left(\frac{8k - 3k}{9}\right)[\text{ Terms inside the bracket are simplified first }]\\[1em] \Rightarrow \frac{6k}{7} - \frac{5k}{9}\\[1em] \Rightarrow \frac{54k - 35k}{63}[\text{ L.C.M. of 7 and 9 = 63 }]\\[1em] \Rightarrow \frac{19k}{63}\end{aligned}$$

Hence, $\bm{\frac{6k}{7} - \left(\frac{8k}{9} - \frac{k}{3}\right) = \frac{19k}{63}}$.

Simplifying,

$$\begin{aligned}\Rightarrow \frac{3a}{8} + \frac{4a}{5} - \left(\frac{a}{2} + \frac{2a}{5}\right)\\[1em] \Rightarrow \frac{3a}{8} + \frac{4a}{5} - \left(\frac{5a + 4a}{10}\right)[\text{ Terms inside the bracket are simplified first }]\\[1em] \Rightarrow \frac{3a}{8} + \frac{4a}{5} - \frac{9a}{10}\\[1em] \Rightarrow \frac{15a + 32a - 36a}{40}[\text{ L.C.M. of 8, 5 and 10 = 40 }]\\[1em] \Rightarrow \frac{11a}{40}\end{aligned}$$

Hence, $\bm{\frac{3a}{8} + \frac{4a}{5} - \left(\frac{a}{2} + \frac{2a}{5}\right) = \frac{11a}{40}}$.

Simplifying,

$$\begin{aligned}\Rightarrow x + \frac{x}{2} + \frac{x}{3}\\[1em] \Rightarrow \frac{6x + 3x + 2x}{6}[\text{ L.C.M. of 1, 2 and 3 = 6 }]\\[1em] \Rightarrow \frac{11x}{6}\end{aligned}$$

Hence, $\bm{x + \frac{x}{2} + \frac{x}{3} = \frac{11x}{6}}$.

Simplifying,

$$\begin{aligned}\Rightarrow \frac{y}{5} + y - \frac{19y}{15}\\[1em] \Rightarrow \frac{3y + 15y - 19y}{15}[\text{ L.C.M. of 5, 1 and 15 = 15 }]\\[1em] \Rightarrow -\frac{y}{15}\end{aligned}$$

Hence, $\bm{\frac{y}{5} + y - \frac{19y}{15} = -\frac{y}{15}}$.

Simplifying,

$$\begin{aligned}\Rightarrow \frac{x}{5} + \frac{x + 1}{2}\\[1em] \Rightarrow \frac{2x + 5(x + 1)}{10}[\text{ L.C.M. of 5 and 2 = 10 }]\\[1em] \Rightarrow \frac{2x + 5x + 5}{10}\\[1em] \Rightarrow \frac{7x + 5}{10}\end{aligned}$$

Hence, $\bm{\frac{x}{5} + \frac{x + 1}{2} = \frac{7x + 5}{10}}$.

Simplifying,

$$\begin{aligned}\Rightarrow x + \frac{x + 2}{3}\\[1em] \Rightarrow \frac{3x + (x + 2)}{3}[\text{ L.C.M. of 1 and 3 = 3 }]\\[1em] \Rightarrow \frac{3x + x + 2}{3}\\[1em] \Rightarrow \frac{4x + 2}{3}\end{aligned}$$

Hence, $\bm{x + \frac{x + 2}{3} = \frac{4x + 2}{3}}$.

Simplifying,

$$\begin{aligned}\Rightarrow \frac{3y}{5} - \frac{y + 2}{2}\\[1em] \Rightarrow \frac{6y - 5(y + 2)}{10}[\text{ L.C.M. of 5 and 2 = 10 }]\\[1em] \Rightarrow \frac{6y - 5y - 10}{10}\\[1em] \Rightarrow \frac{y - 10}{10}\end{aligned}$$

Hence, $\bm{\frac{3y}{5} - \frac{y + 2}{2} = \frac{y - 10}{10}}$.

Simplifying,

$$\begin{aligned}\Rightarrow \frac{2a + 1}{3} + \frac{3a - 1}{2}\\[1em] \Rightarrow \frac{2(2a + 1) + 3(3a - 1)}{6}[\text{ L.C.M. of 3 and 2 = 6 }]\\[1em] \Rightarrow \frac{4a + 2 + 9a - 3}{6}\\[1em] \Rightarrow \frac{13a - 1}{6}\end{aligned}$$

Hence, $\bm{\frac{2a + 1}{3} + \frac{3a - 1}{2} = \frac{13a - 1}{6}}$.

Simplifying,

$$\begin{aligned}\Rightarrow \frac{k + 1}{2} + \frac{2k - 1}{3} - \frac{k + 3}{4}\\[1em] \Rightarrow \frac{6(k + 1) + 4(2k - 1) - 3(k + 3)}{12}[\text{ L.C.M. of 2, 3 and 4 = 12 }]\\[1em] \Rightarrow \frac{6k + 6 + 8k - 4 - 3k - 9}{12}\\[1em] \Rightarrow \frac{11k - 7}{12}\end{aligned}$$

Hence, $\bm{\frac{k + 1}{2} + \frac{2k - 1}{3} - \frac{k + 3}{4} = \frac{11k - 7}{12}}$.

Simplifying,

$$\begin{aligned}\Rightarrow \frac{m}{5} - \frac{m - 2}{3} + m\\[1em] \Rightarrow \frac{3m - 5(m - 2) + 15m}{15}[\text{ L.C.M. of 5, 3 and 1 = 15 }]\\[1em] \Rightarrow \frac{3m - 5m + 10 + 15m}{15}\\[1em] \Rightarrow \frac{13m + 10}{15}\end{aligned}$$

Hence, $\bm{\frac{m}{5} - \frac{m - 2}{3} + m = \frac{13m + 10}{15}}$.

Simplifying,

$$\begin{aligned}\Rightarrow \frac{5(x - 4)}{3} + \frac{2(5x - 3)}{5} + \frac{6(x - 4)}{7}\\[1em] \Rightarrow \frac{175(x - 4) + 42(5x - 3) + 90(x - 4)}{105}[\text{ L.C.M. of 3, 5 and 7 = 105 }]\\[1em] \Rightarrow \frac{175x - 700 + 210x - 126 + 90x - 360}{105}\\[1em] \Rightarrow \frac{475x - 1186}{105}\end{aligned}$$

Hence, $\bm{\frac{5(x - 4)}{3} + \frac{2(5x - 3)}{5} + \frac{6(x - 4)}{7} = \frac{475x - 1186}{105}}$.

Simplifying,

$$\begin{aligned}\Rightarrow \left(p + \frac{p}{3}\right)\left(2p + \frac{p}{2}\right)\left(3p - \frac{2p}{3}\right)\\[1em] \Rightarrow \left(\frac{3p + p}{3}\right)\left(\frac{4p + p}{2}\right)\left(\frac{9p - 2p}{3}\right)\\[1em] \Rightarrow \frac{4p}{3} \times \frac{5p}{2} \times \frac{7p}{3}\\[1em] \Rightarrow \frac{140p^3}{18}\\[1em] \Rightarrow \frac{70p^3}{9}\end{aligned}$$

Hence, $\bm{\left(p + \frac{p}{3}\right)\left(2p + \frac{p}{2}\right)\left(3p - \frac{2p}{3}\right) = \frac{70p^3}{9}}$.

Simplifying,

$$\begin{aligned}\Rightarrow \frac{7}{30} \text{ of } \left(\frac{p}{3} + \frac{7p}{15}\right)\\[1em] \Rightarrow \frac{7}{30} \times \left(\frac{5p + 7p}{15}\right)[\text{ L.C.M. of 3 and 15 = 15 }]\\[1em] \Rightarrow \frac{7}{30} \times \frac{12p}{15}\\[1em] \Rightarrow \frac{84p}{450}\\[1em] \Rightarrow \frac{14p}{75}\end{aligned}$$

Hence, $\bm{\frac{7}{30} \text{ of } \left(\frac{p}{3} + \frac{7p}{15}\right) = \frac{14p}{75}}$.

Simplifying,

$$\begin{aligned}\Rightarrow \left(2p + \frac{p}{7}\right) \div \left(\frac{9p}{10} + 4p\right)\\[1em] \Rightarrow \left(\frac{14p + p}{7}\right) \div \left(\frac{9p + 40p}{10}\right)\\[1em] \Rightarrow \frac{15p}{7} \div \frac{49p}{10}\\[1em] \Rightarrow \frac{15p}{7} \times \frac{10}{49p}\\[1em] \Rightarrow \frac{150}{343}\end{aligned}$$

Hence, $\bm{\left(2p + \frac{p}{7}\right) \div \left(\frac{9p}{10} + 4p\right) = \frac{150}{343}}$.

Simplifying,

$$\begin{aligned}\Rightarrow \left(\frac{5k}{8} - \frac{3k}{5}\right) \div \frac{k}{4}\\[1em] \Rightarrow \left(\frac{25k - 24k}{40}\right) \div \frac{k}{4}[\text{ L.C.M. of 8 and 5 = 40 }]\\[1em] \Rightarrow \frac{k}{40} \times \frac{4}{k}\\[1em] \Rightarrow \frac{1}{10}\end{aligned}$$

Hence, $\bm{\left(\frac{5k}{8} - \frac{3k}{5}\right) \div \frac{k}{4} = \frac{1}{10}}$.

Simplifying,

$$\begin{aligned}\Rightarrow \left(\frac{y}{6} + \frac{2y}{3}\right) \div \left(y + \frac{2y - 1}{3}\right)\\[1em] \Rightarrow \left(\frac{y + 4y}{6}\right) \div \left(\frac{3y + 2y - 1}{3}\right)\\[1em] \Rightarrow \frac{5y}{6} \div \frac{5y - 1}{3}\\[1em] \Rightarrow \frac{5y}{6} \times \frac{3}{5y - 1}\\[1em] \Rightarrow \frac{5y}{2(5y - 1)} = \frac{5y}{10y - 2}\end{aligned}$$

Hence, $\bm{\left(\frac{y}{6} + \frac{2y}{3}\right) \div \left(y + \frac{2y - 1}{3}\right) = \frac{5y}{2(5y - 1)}}$.

$x - y - z = x - (\bm{y + z})$

$x^2 - xy^2 - 2xy - y^2 = x^2 - (\bm{xy^2 + 2xy + y^2})$

$4a - 9 + 2b - 6 = 4a - (\bm{9 - 2b + 6})$

$x^2 - y^2 + z^2 + 3x - 2y = x^2 - (\bm{y^2 - z^2 - 3x + 2y})$

$-2a^2 + 4ab - 6a^2b^2 + 8ab^2 = -2a(\bm{a - 2b + 3ab^2 - 4b^2})$

Solving,

$$\begin{aligned}\Rightarrow 2x - (x + 2y - z)\\[1em] \Rightarrow 2x - x - 2y + z\\[1em] \Rightarrow x - 2y + z\end{aligned}$$

Hence, $\bm{2x - (x + 2y - z) = x - 2y + z}$.

Solving,

$$\begin{aligned}\Rightarrow p + q - (p - q) + (2p - 3q)\\[1em] \Rightarrow p + q - p + q + 2p - 3q\\[1em] \Rightarrow (1 - 1 + 2)p + (1 + 1 - 3)q\\[1em] \Rightarrow 2p - q\end{aligned}$$

Hence, $\bm{p + q - (p - q) + (2p - 3q) = 2p - q}$.

Solving,

$$\begin{aligned}\Rightarrow 9x - (-4x + 5)\\[1em] \Rightarrow 9x + 4x - 5\\[1em] \Rightarrow 13x - 5\end{aligned}$$

Hence, $\bm{9x - (-4x + 5) = 13x - 5}$.

Solving,

$$\begin{aligned}\Rightarrow 6a - (-5a - 8b) + (3a + b)\\[1em] \Rightarrow 6a + 5a + 8b + 3a + b\\[1em] \Rightarrow (6 + 5 + 3)a + (8 + 1)b\\[1em] \Rightarrow 14a + 9b\end{aligned}$$

Hence, $\bm{6a - (-5a - 8b) + (3a + b) = 14a + 9b}$.

Solving,

$$\begin{aligned}\Rightarrow (p - 2q) - (3q - r)\\[1em] \Rightarrow p - 2q - 3q + r\\[1em] \Rightarrow p - 5q + r\end{aligned}$$

Hence, $\bm{(p - 2q) - (3q - r) = p - 5q + r}$.

Solving,

$$\begin{aligned}\Rightarrow 9a(2b - 3a + 7c)\\[1em] \Rightarrow 9a \times 2b - 9a \times 3a + 9a \times 7c\\[1em] \Rightarrow 18ab - 27a^2 + 63ac\end{aligned}$$

Hence, $\bm{9a(2b - 3a + 7c) = 18ab - 27a^2 + 63ac}$.

Solving,

$$\begin{aligned}\Rightarrow -5m(-2m + 3n - 7p)\\[1em] \Rightarrow -5m \times (-2m) - 5m \times 3n - 5m \times (-7p)\\[1em] \Rightarrow 10m^2 - 15mn + 35mp\end{aligned}$$

Hence, $\bm{-5m(-2m + 3n - 7p) = 10m^2 - 15mn + 35mp}$.

Solving,

$$\begin{aligned}\Rightarrow -2x(x + y) + x^2\\[1em] \Rightarrow -2x^2 - 2xy + x^2\\[1em] \Rightarrow -x^2 - 2xy\end{aligned}$$

Hence, $\bm{-2x(x + y) + x^2 = -x^2 - 2xy}$.

Solving,

$$\begin{aligned}\Rightarrow b\left(2b - \frac{1}{b}\right) - 2b\left(b - \frac{1}{b}\right)\\[1em] \Rightarrow 2b^2 - 1 - 2b^2 + 2\\[1em] \Rightarrow 1\end{aligned}$$

Hence, $\bm{b\left(2b - \frac{1}{b}\right) - 2b\left(b - \frac{1}{b}\right) = 1}$.

Solving,

$$\begin{aligned}\Rightarrow 8(2a + 3b - c) - 10(a + 2b + 3c)\\[1em] \Rightarrow 16a + 24b - 8c - 10a - 20b - 30c\\[1em] \Rightarrow (16 - 10)a + (24 - 20)b + (-8 - 30)c\\[1em] \Rightarrow 6a + 4b - 38c\end{aligned}$$

Hence, $\bm{8(2a + 3b - c) - 10(a + 2b + 3c) = 6a + 4b - 38c}$.

Solving,

$$\begin{aligned}\Rightarrow a\left(a + \frac{1}{a}\right) - b\left(b - \frac{1}{b}\right) - c\left(c + \frac{1}{c}\right)\\[1em] \Rightarrow (a^2 + 1) - (b^2 - 1) - (c^2 + 1)\\[1em] \Rightarrow a^2 + 1 - b^2 + 1 - c^2 - 1\\[1em] \Rightarrow a^2 - b^2 - c^2 + 1\end{aligned}$$

Hence, $\bm{a\left(a + \frac{1}{a}\right) - b\left(b - \frac{1}{b}\right) - c\left(c + \frac{1}{c}\right) = a^2 - b^2 - c^2 + 1}$.

Solving,

$$\begin{aligned}\Rightarrow 5x(2x + 3y) - 2x(x - 9y)\\[1em] \Rightarrow 10x^2 + 15xy - 2x^2 + 18xy\\[1em] \Rightarrow 8x^2 + 33xy\end{aligned}$$

Hence, $\bm{5x(2x + 3y) - 2x(x - 9y) = 8x^2 + 33xy}$.

Solving,

$$\begin{aligned}\Rightarrow a + (b + c - d)\\[1em] \Rightarrow a + b + c - d\end{aligned}$$

Hence, $\bm{a + (b + c - d) = a + b + c - d}$.

Solving,

$$\begin{aligned}\Rightarrow 5 - 8x - (6 - x)\\[1em] \Rightarrow 5 - 8x - 6 + x\\[1em] \Rightarrow -1 - 7x\end{aligned}$$

Hence, $\bm{5 - 8x - (6 - x) = -1 - 7x}$.

Solving,

$$\begin{aligned}\Rightarrow 2a + (b - \overline{a - b})\\[1em] \Rightarrow 2a + (b - a + b)\\[1em] \Rightarrow 2a + b - a + b\\[1em] \Rightarrow a + 2b\end{aligned}$$

Hence, $\bm{2a + (b - \overline{a - b}) = a + 2b}$.

Solving,

$$\begin{aligned}\Rightarrow 3x + [4x - (6x - 3)]\\[1em] \Rightarrow 3x + [4x - 6x + 3] \\[1em] \Rightarrow 3x + 4x - 6x + 3 \\[1em] \Rightarrow x + 3\end{aligned}$$

Hence, $\bm{3x + [4x - (6x - 3)] = x + 3}$.

Solving,

$$\begin{aligned}\Rightarrow 5b - \big\lbrace 6a + (8 - b - a)\big\rbrace\\[1em] \Rightarrow 5b - \big\lbrace 6a + 8 - b - a\big\rbrace \\[1em] \Rightarrow 5b - \big\lbrace 5a + 8 - b\big\rbrace\\[1em] \Rightarrow 5b - 5a - 8 + b \\[1em] \Rightarrow 6b - 5a - 8\end{aligned}$$

Hence, $\bm{5b - \big\lbrace6a + (8 - b - a)\big\rbrace = 6b - 5a - 8}$.

Solving,

$$\begin{aligned}\Rightarrow 2x - [5y - (3x - y) + x]\\[1em] \Rightarrow 2x - [5y - 3x + y + x] \\[1em] \Rightarrow 2x - [6y - 2x]\\[1em] \Rightarrow 2x - 6y + 2x \\[1em] \Rightarrow 4x - 6y\end{aligned}$$

Hence, $\bm{2x - [5y - (3x - y) + x] = 4x - 6y}$.

Solving,

$$\begin{aligned}\Rightarrow 6a - 3(a + b - 2)\\[1em] \Rightarrow 6a - 3a - 3b + 6\\[1em] \Rightarrow 3a - 3b + 6\end{aligned}$$

Hence, $\bm{6a - 3(a + b - 2) = 3a - 3b + 6}$.

Solving,

$$\begin{aligned}\Rightarrow 8[m + 2n - p - 7(2m - n + 3p)]\\[1em] \Rightarrow 8[m + 2n - p - 14m + 7n - 21p] \\[1em] \Rightarrow 8[(1 - 14)m + (2 + 7)n + (-1 - 21)p]\\[1em] \Rightarrow 8[-13m + 9n - 22p]\\[1em] \Rightarrow -104m + 72n - 176p\end{aligned}$$

Hence, $\bm{8[m + 2n - p - 7(2m - n + 3p)] = -104m + 72n - 176p}$.

Solving,

$$\begin{aligned}\Rightarrow \big\lbrace 9 - (4p - 6q)\big\rbrace - \big\lbrace 3q - (5p - 10)\big\rbrace \\[1em] \Rightarrow \big\lbrace 9 - 4p + 6q\big\rbrace - \big\lbrace 3q - 5p + 10\big\rbrace \\[1em] \Rightarrow 9 - 4p + 6q - 3q + 5p - 10 \\[1em] \Rightarrow (-4 + 5)p + (6 - 3)q + (9 - 10) \\[1em] \Rightarrow p + 3q - 1\end{aligned}$$

Hence, $\bm{ \big\lbrace 9 - (4p - 6q)\big\rbrace - \big\lbrace 3q - (5p - 10)\big\rbrace = p + 3q - 1}$.

Solving,

$$\begin{aligned}\Rightarrow 2[a - 3 \big\lbrace a + 5(a - 2) + 7\big\rbrace]\\[1em] \Rightarrow 2[a - 3 \big\lbrace a + 5a - 10 + 7\big\rbrace] \\[1em] \Rightarrow 2[a - 3 \big\lbrace 6a - 3\big\rbrace]\\[1em] \Rightarrow 2[a - 18a + 9] \\[1em] \Rightarrow 2[-17a + 9]\\[1em] \Rightarrow -34a + 18\end{aligned}$$

Hence, $\bm{2[a - 3 \big\lbrace a + 5(a - 2) + 7\big\rbrace] = -34a + 18}$.

Solving,

$$\begin{aligned}\Rightarrow 5a - [6a - \big\lbrace 9a - (10a - \overline{4a - 3a})\big\rbrace]\\[1em] \Rightarrow 5a - [6a - \big\lbrace 9a - (10a - a)\big\rbrace] \\[1em] \Rightarrow 5a - [6a - \big\lbrace 9a - 9a\big\rbrace] \\[1em] \Rightarrow 5a - [6a - 0] \\[1em] \Rightarrow 5a - 6a\\[1em] \Rightarrow -a\end{aligned}$$

Hence, $\bm{5a - [6a - \big\lbrace 9a - (10a - \overline{4a - 3a})\big\rbrace] = -a}$.

Solving,

$$\begin{aligned}\Rightarrow 9x + 5 - [4x - \big\lbrace 3x - 2(4x - 3)\big\rbrace]\\[1em] \Rightarrow 9x + 5 - [4x - \big\lbrace 3x - 8x + 6\big\rbrace] \\[1em] \Rightarrow 9x + 5 - [4x - \big\lbrace -5x + 6\big\rbrace]\\[1em] \Rightarrow 9x + 5 - [4x + 5x - 6] \\[1em] \Rightarrow 9x + 5 - [9x - 6]\\[1em] \Rightarrow 9x + 5 - 9x + 6 \\[1em] \Rightarrow 11\end{aligned}$$

Hence, $\bm{9x + 5 - [4x - \big\lbrace 3x - 2(4x - 3)\big\rbrace] = 11}$.

Solving,

$$\begin{aligned}\Rightarrow (x + y - z)x + (z + x - y)y - (x + y - z)z\\[1em] \Rightarrow (x^2 + xy - xz) + (yz + xy - y^2) - (xz + yz - z^2)\\[1em] \Rightarrow x^2 + xy - xz + yz + xy - y^2 - xz - yz + z^2\\[1em] \Rightarrow x^2 - y^2 + z^2 + (1 + 1)xy + (-1 - 1)xz + (1 - 1)yz\\[1em] \Rightarrow x^2 - y^2 + z^2 + 2xy - 2xz\end{aligned}$$

Hence, $\bm{(x + y - z)x + (z + x - y)y - (x + y - z)z = x^2 - y^2 + z^2 + 2xy - 2xz}$.

Solving,

$$\begin{aligned}\Rightarrow -1[a - 3 \big\lbrace b - 4(\overline{a - b} - 8) + 4a\big\rbrace + 10]\\[1em] \Rightarrow -1[a - 3 \big\lbrace b - 4(a - b - 8) + 4a\big\rbrace + 10] \\[1em] \Rightarrow -1[a - 3 \big\lbrace b - 4a + 4b + 32 + 4a\big\rbrace + 10] \\[1em] \Rightarrow -1[a - 3 \big\lbrace 5b + 32\big\rbrace + 10]\\[1em] \Rightarrow -1[a - 15b - 96 + 10] \\[1em] \Rightarrow -1[a - 15b - 86]\\[1em] \Rightarrow -a + 15b + 86\end{aligned}$$

Hence, $\bm{-1[a - 3 \big\lbrace b - 4(\overline{a - b} - 8) + 4a\big\rbrace + 10] = -a + 15b + 86}$.

Solving,

$$\begin{aligned}\Rightarrow p^2 - [x^2 - \big\lbrace x^2 - (q^2 - \overline{x^2 - q^2}) - 2y^2\big\rbrace]\\[1em] \Rightarrow p^2 - [x^2 - \big\lbrace x^2 - (q^2 - x^2 + q^2) - 2y^2\big\rbrace] \\[1em] \Rightarrow p^2 - [x^2 - \big\lbrace x^2 - (2q^2 - x^2) - 2y^2\big\rbrace]\\[1em] \Rightarrow p^2 - [x^2 - \big\lbrace x^2 - 2q^2 + x^2 - 2y^2\big\rbrace] \\[1em] \Rightarrow p^2 - [x^2 - \big\lbrace 2x^2 - 2q^2 - 2y^2\big\rbrace]\\[1em] \Rightarrow p^2 - [x^2 - 2x^2 + 2q^2 + 2y^2] \\[1em] \Rightarrow p^2 - [-x^2 + 2q^2 + 2y^2] \\[1em] \Rightarrow p^2 + x^2 - 2q^2 - 2y^2\end{aligned}$$

Hence, $\bm{p^2 - [x^2 - \big\lbrace x^2 - (q^2 - \overline{x^2 - q^2}) - 2y^2\big\rbrace] = p^2 + x^2 - 2q^2 - 2y^2}$.

Solving,

$$\begin{aligned}\Rightarrow 10 - \big\lbrace 4a - (7 - \overline{a - 5}) - (5a - \overline{1 + a})\big\rbrace\\[1em] \Rightarrow 10 - \big\lbrace 4a - (7 - a + 5) - (5a - 1 - a)\big\rbrace \\[1em] \Rightarrow 10 - \big\lbrace 4a - (12 - a) - (4a - 1)\big\rbrace\\[1em] \Rightarrow 10 - \big\lbrace 4a - 12 + a - 4a + 1\big\rbrace \\[1em] \Rightarrow 10 - \big\lbrace a - 11\big\rbrace\\[1em] \Rightarrow 10 - a + 11 \\[1em] \Rightarrow 21 - a\end{aligned}$$

Hence, $\bm{10 - \big\lbrace 4a - (7 - \overline{a - 5}) - (5a - \overline{1 + a})\big\rbrace = 21 - a}$.

Solving,

$$\begin{aligned}\Rightarrow 7a - [8a - \big\lbrace 11a - (12a - \overline{6a - 5a})\big\rbrace]\\[1em] \Rightarrow 7a - [8a - \big\lbrace 11a - (12a - a)\big\rbrace] \\[1em] \Rightarrow 7a - [8a - \big\lbrace 11a - 11a\big\rbrace] \\[1em] \Rightarrow 7a - [8a - 0] \\[1em] \Rightarrow 7a - 8a\\[1em] \Rightarrow -a\end{aligned}$$

Hence, $\bm{7a - [8a - \big\lbrace 11a - (12a - \overline{6a - 5a})\big\rbrace] = -a}$.

Solving,

$$\begin{aligned}\Rightarrow 8x - [4y - \big\lbrace 4x + (2x - \overline{2y - 2x})\big\rbrace]\\[1em] \Rightarrow 8x - [4y - \big\lbrace 4x + (2x - 2y + 2x)\big\rbrace] \\[1em] \Rightarrow 8x - [4y - \big\lbrace 4x + (4x - 2y)\big\rbrace]\\[1em] \Rightarrow 8x - [4y - \big\lbrace 4x + 4x - 2y\big\rbrace] \\[1em] \Rightarrow 8x - [4y - \big\lbrace 8x - 2y\big\rbrace]\\[1em] \Rightarrow 8x - [4y - 8x + 2y] \\[1em] \Rightarrow 8x - [6y - 8x]\\[1em] \Rightarrow 8x - 6y + 8x \\[1em] \Rightarrow 16x - 6y\end{aligned}$$

Hence, $\bm{8x - [4y - \big\lbrace 4x + (2x - \overline{2y - 2x})\big\rbrace] = 16x - 6y}$.

Solving,

$$\begin{aligned}\Rightarrow x - (3y - \overline{4z - 3x} + 2z - \overline{5y - 7x})\\[1em] \Rightarrow x - (3y - 4z + 3x + 2z - 5y + 7x) \\[1em] \Rightarrow x - [(3 + 7)x + (3 - 5)y + (-4 + 2)z]\\[1em] \Rightarrow x - (10x - 2y - 2z)\\[1em] \Rightarrow x - 10x + 2y + 2z \\[1em] \Rightarrow -9x + 2y + 2z\end{aligned}$$

Hence, $\bm{x - (3y - \overline{4z - 3x} + 2z - \overline{5y - 7x}) = -9x + 2y + 2z}$.

$\frac{12}{x^2} = 12x^{-2} \text{ and } \frac{5}{x} = 5x^{-1}$ have negative exponents of $x$, so those two expressions are not polynomials. In $8 - 5x$ and 8.7, the exponents of $x$ are whole numbers.

Hence, option 3 is the correct option.

Substituting $x = 2, y = 3$ and $z = 5$,

$$\begin{aligned}\Rightarrow 3y(x^3 - z) \\[1em] \Rightarrow 3(3)[(2)^3 - 5] \\[1em] \Rightarrow 9(8 - 5) = 9 \times 3 = 27.\end{aligned}$$

Hence, option 2 is the correct option.

Solving,

$$\begin{aligned}\Rightarrow (4x + 8x^2 + 1) + (13 - x^2) + (6x - 10 - 8x^2)\\[1em] \Rightarrow (8 - 1 - 8)x^2 + (4 + 6)x + (1 + 13 - 10)\\[1em] \Rightarrow -x^2 + 10x + 4.\end{aligned}$$

Hence, option 1 is the correct option.

Solving,

$$\begin{aligned}\Rightarrow (3x^2 + 7 - 6x) - (3x - 4x^2 - 15)\\[1em] \Rightarrow 3x^2 + 7 - 6x - 3x + 4x^2 + 15\\[1em] \Rightarrow 7x^2 - 9x + 22.\end{aligned}$$

Hence, option 2 is the correct option.

Solving,

$$\begin{aligned}\Rightarrow (5x^3y^2z^5) \times (-4x^2y^3)\\[1em] \Rightarrow (5 \times -4) \times x^{3+2} \times y^{2+3} \times z^5\\[1em] \Rightarrow -20x^5y^5z^5.\end{aligned}$$

Hence, option 3 is the correct option.

Solving,

$$\begin{aligned}\Rightarrow (3x - 4y)(5x - 6y)\\[1em] \Rightarrow 15x^2 - 18xy - 20xy + 24y^2\\[1em] \Rightarrow 15x^2 - 38xy + 24y^2.\end{aligned}$$

Hence, option 4 is the correct option.

Solving,

$$\begin{aligned}\Rightarrow (x - 1)(x^2 + x + 1)\\[1em] \Rightarrow x^3 + x^2 + x - x^2 - x - 1\\[1em] \Rightarrow x^3 - 1.\end{aligned}$$

Hence, option 1 is the correct option.

Solving,

$$\begin{aligned}\Rightarrow (x + 1)(x^2 - x + 1)\\[1em] \Rightarrow x^3 - x^2 + x + x^2 - x + 1\\[1em] \Rightarrow x^3 + 1.\end{aligned}$$

Hence, option 2 is the correct option.

Solving,

$$\begin{aligned}\Rightarrow \frac{-63x^5y^3z^{10}}{-7x^2y^5z^7} \\[1em] \Rightarrow 9 \times x^{5-2} \times y^{3-5} \times z^{10-7} \\[1em] \Rightarrow \frac{9x^3z^3}{y^2}\end{aligned}$$

Hence, option 3 is the correct option.

Solving,

$$\begin{aligned}\Rightarrow \frac{4x^3y - 12x^2y^2}{-4xy} \\[1em] \Rightarrow \frac{4x^3y}{-4xy} - \frac{12x^2y^2}{-4xy} \\[1em] \Rightarrow -x^2 + 3xy\end{aligned}$$

Hence, option 3 is the correct option.

Solving,

$$\begin{aligned}\Rightarrow 2x - 2(x + 3) + 4(x - 1)\\[1em] \Rightarrow 2x - 2x - 6 + 4x - 4\\[1em] \Rightarrow 4x - 10.\end{aligned}$$

Hence, option 4 is the correct option.

Solving,

$$\begin{aligned}\Rightarrow (8 - x)(3 - 2x) \\[1em] \Rightarrow 24 - 16x - 3x + 2x^2 \\[1em] \Rightarrow 2x^2 - 19x + 24.\end{aligned}$$

The highest power of $x$ is 2, so the degree is 2.

Hence, option 1 is the correct option.

Only the term $3xy^2$ contains $x$, and $3xy^2 = (3y^2) \times x$. So the coefficient of $x$ is $3y^2$.

Hence, option 3 is the correct option.

Solving,

$$\begin{aligned}\Rightarrow (9 + x + 2x^2) - (3x^2 - 5x) - (3 + 2x - x^2)\\[1em] \Rightarrow 9 + x + 2x^2 - 3x^2 + 5x - 3 - 2x + x^2\\[1em] \Rightarrow (2 - 3 + 1)x^2 + (1 + 5 - 2)x + (9 - 3)\\[1em] \Rightarrow 6 + 4x.\end{aligned}$$

Hence, option 1 is the correct option.

Substituting $x = 2$,

$3(4x - 1) - 2(5 - 2x) = 3(8 - 1) - 2(5 - 4) = 3(7) - 2(1) = 21 - 2 = 19$.

Hence, option 2 is the correct option.

$\frac{4}{x} = 4x^{-1}$ has degree -1, which is not a whole number. So the given expression is not a polynomial and Statement 1 is false.

Statement 2 is true, because an algebraic expression is a polynomial only when the exponent of the variable in each of its terms is a whole number (0, 1, 2, 3, .....), i.e. the degree of every term is a whole number. A term with a negative or fractional exponent, such as $\frac{4}{x} = 4x^{-1}\text{ or }\sqrt{x} = x^{\frac{1}{2}}$, is not allowed in a polynomial.

Hence, option 4 is the correct option.

Since $3x^0 = 3$ (as $x \neq 0$), the given expression does have a constant term 3. So, Statement 1 is false.

Statement 2 is true, because in an algebraic expression having only one variable, a term whose value does not change with the value of the variable, i.e. a term free of the variable, is called the constant term. Here $3x^0 = 3$ is free of $x$ and so 3 is the constant term of the given expression.

Hence, option 4 is the correct option.

In the term $\frac{1}{2}x^{\frac{5}{2}}\sqrt{y} = \frac{1}{2}x^{\frac{5}{2}}y^{\frac{1}{2}}$, the power of $y$ is not a whole number. So the given expression is not a polynomial and hence its degree cannot be stated as 3. Thus, Assertion (A) is false.

Reason (R) is true, because the degree of a term of a polynomial is the sum of the exponents of the variables occurring in that term, and the degree of the polynomial is the greatest of the degrees of its terms. For example, in $4x^2y^3 + 7xy$, the degrees of the terms are (2 + 3) = 5 and (1 + 1) = 2, so the degree of the polynomial is 5.

Hence, option 2 is the correct option.

In $3xyz + 2x^2y + z^2x$, the degree of each term is (1 + 1 + 1) = 3, (2 + 1) = 3 and (2 + 1) = 3. So it is a polynomial of degree 3 (a cubic polynomial) in the three variables $x, y$ and $z$. Thus, Assertion (A) is true.

A polynomial of degree 3 can also be in a single variable, for example $x^3 + 2x + 1$. So Reason (R) is also true.

Hence, option 3 is the correct option.

In $\frac{x}{2} + \frac{y^2}{5} + \frac{yz}{7}$, the degrees of the terms are 1, 2 and 2, so its degree is 2.

In $\frac{xy}{2} + \frac{z^2}{5} + \frac{xz}{7}$, the degrees of the terms are 2, 2 and 2, so its degree is 2.

Thus, Assertion (A) is true. A polynomial of degree 2 is called a quadratic polynomial, so Reason (R) is also true.

Hence, option 3 is the correct option.