Expansions

(2x + 7y)² = (2x)² + 2(2x)(7y) + (7y)²

= 4x² + 28xy + 49y²

$$\begin{aligned}\Big(\frac{1}{2}x + \frac{2}{3}y\Big)^2 = \Big(\frac{1}{2}x\Big)^2 + 2\Big(\frac{1}{2}x\Big)\Big(\frac{2}{3}y\Big) + \Big(\frac{2}{3}y\Big)^2 \\[1em] = \frac{1}{4}x^2 + \frac{2}{3}xy + \frac{4}{9}y^2\end{aligned}$$

$$\begin{aligned}\Big(3x + \frac{1}{2x} \Big)^2 = (3x)^2 + 2 (3x)\Big(\frac{1}{2x}\Big) + \Big(\frac{1}{2x}\Big)^2 \\[1em] = 9x^2 + 3 + \Big(\frac{1}{4x^2}\Big)\end{aligned}$$

(3x²y + 5z)² = (3x²y)² + 2 (3x²y)(5z) + (5z)²

= 9x⁴y² + 30x²yz + 25z²

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

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

We know that (x+a)(x+b) = x² + (a + b)x + ab

∴ (x+3)(x+5) = x² + (3 + 5)x + (3)(5) = x² + 8x + 15

We know that (x+a)(x-b) = x² + (a - b)x - ab

∴ (x+3)(x-5) = x² + (3 - 5)x - (3)(5) = x² - 2x - 15

We know that (x-a)(x+b) = x² - (a - b)x - ab

∴ (x-7)(x+9) = x² - (7 - 9)x - (7)(9) = x² + 2x - 63

We know that (x-a)(x-b) = x² - (a + b)x + ab

∴ (x-2y)(x-3y) = x² - (2y + 3y)x + (2y)(3y)

= x² - 5xy + 6y²

(x - 2y - z)² = [(x) + (-2y) + (-z)]²

= (x)² + (-2y)² + (-z)² + 2 [(x)(-2y) + (-2y)(-z) + (-z)(x)]

= x² + 4y² + z² + 2[-2xy + 2yz - xz]

= x² + 4y² + z² - 4xy + 4yz - 2xz

(2x - 3y + 4z)² = [(2x) + (-3y) + (4z)]²

= (2x)² + (-3y)² + (4z)² + 2 [(2x)(-3y) + (-3y)(4z) + (4z)(2x)]

= 4x² + 9y² + 16z² + 2[-6xy - 12yz + 8xz]

= 4x² + 9y² + 16z² - 12xy - 24yz + 16xz

$$\begin{aligned}\Big(2x + \frac{3}{x} - 1\Big)^2 = \Big[\Big(2x\Big) + \Big(\frac{3}{x}\Big) + \Big(-1\Big)\Big]^2 \\[1em] = (2x)^2 + \Big(\frac{3}{x}\Big)^2 + (-1)^2 + 2 \Big[\Big(2x\Big)\Big(\frac{3}{x}\Big) + \Big(\frac{3}{x}\Big)\Big(-1\Big)+\Big(-1\Big)\Big(2x\Big) \Big] \\[1em] = 4x^2 + \frac{9}{x^2} + 1 + 2\Big[6 - \frac{3}{x} - 2x \Big] \\[1em] = 4x^2 + \frac{9}{x^2} + 1 + 12 - \frac{6}{x} - 4x \\[1em] = 4x^2 + \frac{9}{x^2} + 13 - \frac{6}{x} - 4x \\[1em]\end{aligned}$$

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

(x+2)³ = x³ + (2)³ + 3(x)(2)(x+2)

= x³ + 8 + 6x² + 12x

(2a+b)³ = (2a)³ + (b)³ + 3(2a)(b)(2a+b)

= 8a³ + b³ + 12a²b + 6ab²

$$\begin{aligned}\Big(3x + \frac{1}{x}\Big)^3 = (3x)^3 + \Big(\frac{1}{x}\Big)^3 + 3(3x)\Big(\frac{1}{x}\Big)\Big(3x + \frac{1}{x}\Big) \\[1em] = 27x^3 + \frac{1}{x^3} + 27x + \frac{9}{x}\end{aligned}$$

(2x-1)³ = (2x)³ - (1)³ - 3(2x)(1)(2x-1)

= 8x³ - 1 -12x² + 6x

(5x-3y)³ = (5x)³ - (3y)³ - 3(5x)(3y)(5x-3y)

= 125x³ - 27y³ - 225x²y + 135xy²

$$\begin{aligned}\Big(2x - \frac{1}{3y}\Big)^3 = (2x)^3 - \Big(\frac{1}{3y}\Big)^3 - 3(2x)\Big(\frac{1}{3y}\Big)\Big(2x - \frac{1}{3y}\Big) \\[1em] = 8x^3 - \frac{1}{27y^3} - \frac{4x^2}{y} + \frac{2x}{3y^2}\end{aligned}$$

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

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

(3x-1)² - (3x-2)(3x+1) = [(3x-1)²] - [(3x-2)(3x+1)]

= [(3x)² - 2(3x)(1) + (1)²] - [(3x)² + 3x - 6x - 2]

= (9x² - 6x + 1) - (9x² - 3x - 2)

= 9x² - 6x + 1 - 9x² + 3x + 2

= -3x + 3

= 3 - 3x

(4x+3y)² - (4x-3y)² - 48xy = [(4x)² + 2(4x)(3y) + (3y)²] - [(4x)² - 2(4x)(3y) + (3y)²] - 48xy

= [16x² + 24xy + 9y²] - [16x² - 24xy + 9y²] - 48xy

= 16x² + 24xy + 9y² - 16x² + 24xy - 9y² - 48xy

= 48xy - 48xy = 0

(7p+9q)(7p-9q) = (7p)² - (9q)²

= 49p² - 81q²

$\Big(2x - \frac{3}{x}\Big)\Big(2x + \frac{3}{x}\Big) = \Big(2x\Big)^2 - \Big(\frac{3}{x}\Big)^2 = 4x^2 - \frac{9}{x^2}$

(2x - y + 3)(2x - y - 3)

Let (2x - y) = a

Then, the given expression = (a+3)(a-3) = (a)² - (3)²

= a² - 9 = (2x-y)² - 9 = (2x)² - 2(2x)(y) + y² - 9

= 4x² - 4xy + y² - 9

(3x+y-5)(3x-y-5) = (3x-5+y)(3x-5-y)

Let (3x-5)=a

Then, the given expression = (a+y)(a-y)

= a² - y² = (3x-5)² - y²

= [(3x)² - 2(3x)(5) + (5)²] - y²

= 9x² - 30x + 25 - y²

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

Let (x - 3) = a

Then, the given expression = $$\begin{aligned}\Big(a+\frac{2}{x}\Big)\Big(a-\frac{2}{x}\Big) = \Big(a\Big)^2 - \Big(\frac{2}{x}\Big)^2 \\[1em] = a^2 - \frac{4}{x^2} = (x-3)^2 - \frac{4}{x^2} = x^2 -2(x)(3) + 3^2 - \frac{4}{x^2} \\[1em] x^2 - 6x + 9 - \frac{4}{x^2}\end{aligned}$$

(5-2x)(5+2x)(25+4x²) = [(5)² - (2x)²] (25+4x²)

= (25-4x²)(25+4x²)

= (25)² - (4x²)² = 625 - 16x⁴

(x+2y+3)(x+2y+7)

Let (x+2y)=z

Then the given expression = (z+3)(z+7)

We know that (x+a)(x+b) = x² + (a + b)x + ab

∴ (z+3)(z+7) = z² + (3+7)z + (3)(7)

= z² + 10z + 21

= (x+2y)² + 10(x+2y) + 21

= x² + 2(x)(2y) + (2y)² + 10x + 20y + 21

= x² + 4xy + 4y² + 10x + 20y + 21

= x² + 10x + 4y² + 20y + 4xy + 21

(2x+y+5)(2x+y-9)

Let (2x+y)=z

Then the given expression = (z+5)(z-9)

We know that (x+a)(x-b) = x² + (a - b)x - ab

∴ (z+5)(z-9) = z² + (5-9)z - (5)(9)

= z² - 4z - 45

= (2x+y)² - 4(2x+y) - 45

= (2x)² + 2(2x)(y) + (y)² - 8x - 4y - 45

= 4x² + 4xy + y² - 8x - 4y - 45

= 4x² - 8x + y² - 4y + 4xy - 45

Let (x - 2y) = z

Then the given expression = (z - 5)(z + 3)

We know that (x - a)(x + b) = x² - (a - b)x - ab

∴ (z - 5)(z + 3) = z² - (5 - 3)z - (5)(3)

= z² - 2z - 15

= (x - 2y)² - 2(x - 2y) - 15

= (x)² - 2(x)(2y) + (2y)² - 2x + 4y - 15

= x² - 4xy + 4y² - 2x + 4y - 15

Let (3x - 4y) = z

Then the given expression = (z - 2)(z - 6)

We know that (x - a)(x - b) = x² - (a + b)x + ab

∴ (z - 2)(z - 6) = z² - (2 + 6)z + (2)(6)

= z² - 8z + 12

= (3x - 4y)² - 8(3x - 4y) + 12

= (3x)² - 2(3x)(4y) + (4y)² - 24x + 32y + 12

= 9x² - 24xy + 16y² - 24x + 32y + 12

(2p + 3q)(4p² - 6pq + 9q²) = (2p + 3q)[(2p)² - (2p)(3q) + (3q)²]

We know that (a + b)(a² - ab + b²) = a³ + b³

∴ (2p + 3q)[(2p)² - (2p)(3q) + (3q)²] = (2p)³ + (3q)³

= 8p³ + 27q³

We know that (a + b)(a² - ab + b²) = a³ + b³

∴ Given Expression = $\Big(x + \frac{1}{x}\Big)\Big(x^2 - 1 + \frac{1}{x^2}\Big)$

$= x^3 + \frac{1}{x^3}$

(3p - 4q)(9p² +12pq + 16q²) = (3p - 4q)[(3p)² + (3p)(4q) + (4q)²]

We know that (a - b)(a² + ab + b²) = a³ - b³

∴ (3p - 4q)[(3p)² + (3p)(4q) + (4q)²] = (3p)³ - (4q)³

= 27p³ - 64q³

We know that (a - b)(a² + ab + b²) = a³ - b³

∴ Given Expression = $\Big(x - \frac{3}{x}\Big)\Big(x^2 + 3 + \frac{9}{x^2}\Big)$

$$\begin{aligned}= x^3 - \Big(\frac{3}{x}\Big)^3 \\[1em] = x^3 - \frac{27}{x^3}\end{aligned}$$

We know that (a + b + c)(a² + b² + c² - ab - bc - ca) = a³ + b³ + c³ - 3abc

∴ The Given Expression = (2x + 3y + 4z)[(2x)² + (3y)² + (4z)² - (2x)(3y) - (3y)(4z) - (4z)(2x)]

= (2x)³ + (3y)³ + (4z)³ - 3(2x)(3y)(4z)

= 8x³ + 27y³ + 64z³ - 72xyz

We know that (x + a)(x + b)(x + c) = x³ + (a + b + c)x² + (ab + bc + ca)x + abc

∴ (x + 1)(x + 2)(x + 3) = x³ + (1 + 2 + 3)x² + [(1)(2) + (2)(3) + (3)(1)]x + (1)(2)(3)

= x³ + 6x² + (2 + 6 + 3)x + 6

= x³ + 6x² + 11x + 6

We know that (x + a)(x + b)(x + c) = x³ + (a + b + c)x² + (ab + bc + ca)x + abc

∴ (x - 2)(x - 3)(x + 4) = x³ + [(-2) + (-3) + 4]x² + [(-2)(-3) + (-3)(4) + (4)(-2)]x + (-2)(-3)(4)

= x³ - x² + (6 -12 - 8)x + 24

= x³ - x² - 14x + 24

We know that (x + a)(x + b)(x + c) = x³ + (a + b + c)x² + (ab + bc + ca)x + abc

Comparing (x - 3)(x + 7)(x - 4) with (x + a)(x + b)(x + c).
Here a = -3, b = 7, c = -4.

∴ Coefficient of x² = a + b + c = (-3) + 7 + (-4) = 0 and
    coefficient of x = ab + bc + ca

    = (-3) x 7 + 7 x (-4) + (-4) x (-3)

    = -21 - 28 + 12 = -37

Given,
    a² + 4a + x = (a + 2)²

⇒ a² + 4a + x = a² + 2(a)(2) + 2²

⇒ a² - a² +4a - 4a + x = 4

⇒ x = 4

(101)² = (100 + 1)²

Using (a + b)² = a² + 2ab + b²,

(100 + 1)² = (100)² + 2(100)(1) + (1)²

= 10000 + 200 + 1 = 10201

(1003)² = (1000 + 3)²

Using (a + b)² = a² + 2ab + b²,

(1000 + 3)² = (1000)² + 2(1000)(3) + (3)²

= 1000000 + 6000 + 9 = 1006009

(10.2)² = (10 + 0.2)²

Using (a + b)² = a² + 2ab + b²,

(10 + 0.2)² = (10)² + 2(10)(0.2) + (0.2)²

= 100 + 4 + 0.04 = 104.04

(99)² = (100 - 1)²

Using (a - b)² = a² - 2ab + b²,

(100 - 1)² = (100)² - 2(100)(1) + (1)²

= 10000 - 200 + 1 = 9801

(997)² = (1000 - 3)²

Using (a - b)² = a² - 2ab + b²,

(1000 - 3)² = (1000)² - 2(1000)(3) + (3)²

= 1000000 - 6000 + 9 = 994009

(9.8)² = (10 - 0.2)²

Using (a - b)² = a² - 2ab + b²,

(10 - 0.2)² = (10)² - 2(10)(0.2) + (0.2)²

= 100 - 4 + 0.04 = 96.04

(103)³ = (100 + 3)³

Using (a + b)³ = a³ + b³ + 3(a)(b)(a + b),

(100 + 3)³ = (100)³ + (3)³ + 3(100)(3)(100 + 3)

= 1000000 + 27 + 900(103)

= 1000000 + 27 + 92700

= 1092727

(99)³ = (100 - 1)³

Using (a - b)³ = a³ - b³ - 3(a)(b)(a - b),

(100 - 1)³ = (100)³ - (1)³ - 3(100)(1)(100 - 1)

= 1000000 - 1 - 300(99)

= 1000000 - 1 - 29700

= 970299

(10.1)³ = (10 + 0.1)³

Using (a + b)³ = a³ + b³ + 3(a)(b)(a + b),

(10 + 0.1)³ = (10)³ + (0.1)³ + 3(10)(0.1)(10.1)

= 1000 + 0.001 + 30.3

= 1030.301

Given,
     2a - b + c = 0
⇒ (2a + c) = b

On squaring both sides,

    (2a + c)² = b²

⇒ 4a² + c² + 2(2a)(c) = b²

⇒ 4a² - b² + c² + 4ac = 0

Hence proved.

We know that if a + b + c = 0 then a³ + b³ + c³ = 3abc

Given,

a + b + 2c = 0

∴ (a)³ + (b)³ + (2c)³ = 3(a)(b)(2c)

⇒ a³ + b³ + 8c³ = 6abc

Hence Proved.

Given,

x + 2y - 3 = 0

⇒ x + 2y = 3

Simplifying,

⇒ x³ + 8y³ + 6x²y + 12xy² - 125

⇒ x³ + (2y)³ + 6xy(x + 2y) - 125

⇒ x³ + (2y)³ + 3.x.2y.(x + 2y) - 125

⇒ (x + 2y)³ - 125

⇒ 3³ - 125

⇒ 27 - 125

⇒ -98.

Hence, the value of x³ + 8y³ + 6x²y + 12xy² - 125 = -98.

We know if a + b + c = 0, a³ + b³ + c³ = 3abc.

On dividing each side by abc,
$$\begin{aligned}\phantom{\Rightarrow} \frac{a^3}{abc} + \frac{b^3}{abc} + \frac{c^3}{abc} = \frac{3abc}{abc} \\[1em] \Rightarrow \frac{a^2}{bc} + \frac{b^2}{ca} + \frac{c^2}{ab} = 3\end{aligned}$$

∴ Value of $\bold{\frac{a^2}{bc} + \frac{b^2}{ca} + \frac{c^2}{ab} = 3}$

Using (x + y)³ = x³ + y³ + 3(x)(y)(x + y)

Putting x + y = 4 in above:

    x³ + y³ + 3xy(4) = (4)³

⇒ x³ + y³ + 12xy = 64

⇒ x³ + y³ + 12xy - 64 = 0

∴ Value of x³ + y³ + 12xy - 64 = 0

(i) Let a = 27, b = -17, c = -10

Then,

a + b + c = 37 - 17 - 10 = 0

Thus, a³ + b³ + c³ = 3abc

∴ (27)³ + (-17)³ + (-10)³ = 3(27)(-17)(-10) = 13770

(ii) Let a = -28, b = 15, c = 13

Then,

a + b + c = -28 + 15 + 13 = 0

Thus, a³ + b³ + c³ = 3abc

∴ (-28)³ + (15)³ + (13)³ = 3(-28)(15)(13) = -16380

Let x = 86 and y = 14.

Hence, above equation can be written as,

$$\begin{aligned}\Rightarrow \frac{x^3 + y^3}{x^2 - xy + y^2} \\[1em] = \frac{(x + y)(x^2 - xy + y^2)}{(x^2 - xy + y^2)} \\[1em] = x + y \\[1em] = 86 + 14 \\[1em] = 100.\end{aligned}$$

Hence, value of $\frac{86 \times 86 \times 86 + 14 \times 14 \times 14}{86 \times 86 - 86 \times 14 + 14 \times 14}$ = 100.

We know that,

(x - y)² = x² - 2xy + y²

⇒ x² + y² = (x - y)² + 2xy.

Substituting values we get,

⇒ x² + y² = (8)² + 2 × 5
⇒ x² + y² = 64 + 10
⇒ x² + y² = 74.

Hence, x² + y² = 74.

We know that,

(x + y)² = x² + 2xy + y²

⇒ x² + y² = (x + y)² - 2xy.

⇒ 2(x² + y²) = 2[(x + y)² - 2xy]

Substituting values we get,

⇒ 2(x² + y²) = 2[(10)² - 2 × 21]

⇒ 2(x² + y²) = 2(100 - 42)

⇒ 2(x² + y²) = 2 × 58

⇒ 2(x² + y²) = 116.

Hence, 2(x² + y²) = 116.

We know that,

a² + b² = (a + b)² - 2ab.

∴ 4a² + 9b² = (2a)² + (3b)² = (2a + 3b)² - 12ab.

Substituting values we get,

⇒ 4a² + 9b² = (7)² - 12 × 2

⇒ 4a² + 9b² = 49 - 24

⇒ 4a² + 9b² = 25.

Hence, 4a² + 9b² = 25.

We know that,

a² + b² = (a - b)² + 2ab.

9x² + 16y² = (3x)² + (4y)² = (3x - 4y)² + 24xy.

Substituting values we get,

⇒ 9x² + 16y² = (16)² + 24 × 4

⇒ 9x² + 16y² = 256 + 96

⇒ 9x² + 16y² = 352.

Hence, 9x² + 16y² = 352.

We know that,

(x + y)² = x² + y² + 2xy .....(i)

(x - y)² = x² + y² - 2xy ....(ii)

Adding eqn. (i) and (ii) we get,

(x + y)² + (x - y)² = x² + x² + y² + y² + 2xy - 2xy

= 2x² + 2y².

∴ 2x² + 2y² = (x + y)² + (x - y)².

Substituting values we get,

⇒ 2x² + 2y² = (8)² + (2)²

⇒ 2x² + 2y² = 64 + 4 = 68.

Hence, 2x² + 2y² = 68.

(i) We know that,

(a + b)² = a² + b² + 2ab

∴ (a + b) = $\sqrt{a^2 + b^2 + 2ab}$

Substituting values we get,

$$\begin{aligned}\Rightarrow (a + b) = \sqrt{13 + 2 \times 6} \\[1em] \Rightarrow (a + b) = \sqrt{13 + 12} \\[1em] \Rightarrow (a + b) = \sqrt{25} \\[1em] \Rightarrow (a + b) = \pm 5.\end{aligned}$$

Hence, a + b = ±5.

(ii) We know that,

(a - b)² = a² + b² - 2ab

∴ (a - b) = $\sqrt{a^2 + b^2 - 2ab}$

Substituting values we get,

$$\begin{aligned}\Rightarrow (a - b) = \sqrt{13 - 2 \times 6} \\[1em] \Rightarrow (a - b) = \sqrt{13 - 12} \\[1em] \Rightarrow (a - b) = \sqrt{1} \\[1em] \Rightarrow (a - b) = \pm 1.\end{aligned}$$

Hence, a - b = ±1.

(i) We know that,

(a - b)² = a² + b² - 2ab

(a - b)² = a² + b² + 2ab -2ab - 2ab

(a - b)² = (a + b)² - 4ab

(a - b) = $\sqrt{(a + b)^2 - 4ab}$

Substituting values we get,

$$\begin{aligned}\Rightarrow (a - b) = \sqrt{(4)^2 - 4 \times (-12)} \\[1em] \Rightarrow (a - b) = \sqrt{16 + 48} \\[1em] \Rightarrow (a - b) = \sqrt{64} \\[1em] \Rightarrow (a - b) = \pm 8.\end{aligned}$$

Hence, a - b = ±8.

(ii) We know that,

a² - b² = (a + b)(a - b).

Substituting values we get,

⇒ a² - b² = 4 × ±8 = ±32.

Hence, a² - b² = ±32.

(i) We know that,

(p - q)² = p² + q² - 2pq

⇒ (p - q)² = p² + q² + 2pq - 2pq - 2pq

⇒ (p - q)² = (p + q)² - 4pq

⇒ (p + q)² = (p - q)² + 4pq

⇒ (p + q) = $\sqrt{(p - q)^2 + 4pq}$

Substituting value we get,

$$\begin{aligned}\Rightarrow p + q = \sqrt{9^2 + 4 \times 36} \\[1em] \Rightarrow p + q = \sqrt{81 + 144} \\[1em] \Rightarrow p + q = \sqrt{225} \\[1em] \Rightarrow p + q = \pm 15.\end{aligned}$$

Hence, p + q = ±15.

(ii) p² - q² = (p - q)(p + q).

Substituting value we get,

⇒ p² - q² = 9 × ±15 = ±135.

Hence, p² - q² = ±135.

(i) We know that,

(x + y)² = x² + y² + 2xy .....(i)

(x - y)² = x² + y² - 2xy ....(ii)

Adding eqn. (i) and (ii) we get,

(x + y)² + (x - y)² = x² + x² + y² + y² + 2xy - 2xy = 2x² + 2y².

⇒ 2x² + 2y² = (x + y)² + (x - y)².

∴ x² + y² = $\frac{(x + y)^2 + (x - y)^2}{2}$

Substituting values we get,

$$\begin{aligned}\Rightarrow x^2 + y^2 = \frac{(6)^2 + (4)^2}{2} \\[1em] = \frac{36 + 16}{2} \\[1em] = \frac{52}{2} \\[1em] = 26.\end{aligned}$$

Hence, x² + y² = 26.

(ii) We know that,

(x + y)² = x² + y² + 2xy .....(i)

(x - y)² = x² + y² - 2xy ....(ii)

Subtracting eqn. (ii) from (i) we get,

(x + y)² - (x - y)² = x² - x² + y² - y² + 2xy - (-2xy) = 4xy.

⇒ (x + y)² - (x - y)² = 4xy.

∴ xy = $\frac{(x + y)^2 - (x - y)^2}{4}$

Substituting values we get,

$$\begin{aligned}xy = \frac{6^2 - 4^2}{4} \\[1em] = \frac{36 - 16}{4} \\[1em] = \frac{20}{4} \\[1em] = 5.\end{aligned}$$

Hence, xy = 5.

Given,

$$\begin{aligned}\phantom{\therefore} x - 3 = \frac{1}{x} \\[1em] \therefore x - \frac{1}{x} = 3 \\[1em]\end{aligned}$$

We know that,

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

Substituting values we get,

$$\begin{aligned}x^2 + \frac{1}{x^2} = (3)^2 - 2 \\[1em] = 9 - 2 \\[1em] = 7.\end{aligned}$$

Hence, $x^2 + \frac{1}{x^2}$ = 7.

(i) We know that,

(x + y)² = x² + y² + 2xy .....(i)

(x - y)² = x² + y² - 2xy .....(ii)

Subtracting eqn. (ii) from (i) we get,

(x + y)² - (x - y)² = x² - x² + y² - y² + 2xy - (-2xy) = 4xy.

⇒ (x + y)² - (x - y)² = 4xy.

∴ (x - y) = $\sqrt{(x + y)^2 - 4xy}$.

Substituting values we get,

$$\begin{aligned}x - y = \sqrt{8^2 - 4 \times 3\frac{3}{4}} \\[1em] = \sqrt{64 - 4 \times \frac{15}{4}} \\[1em] = \sqrt{64 - 15} \\[1em] = \sqrt{49} \\[1em] = \pm 7.\end{aligned}$$

Hence, x - y = ±7.

(ii) We know that,

3(x² + y²) = 3[(x + y)² - 2xy].

Substituting values we get,

$$\begin{aligned}3(x^2 + y^2) = 3\Big[(8)^2 - 2 \times 3\frac{3}{4}\Big] \\[1em] = 3\Big(64 - 2 \times \frac{15}{4}\Big) \\[1em] = 3\Big(64 - \frac{15}{2}\Big) \\[1em] = 3\Big(\frac{128 - 15}{2}\Big) \\[1em] = 3 \times \frac{113}{2} \\[1em] = \frac{339}{2} \\[1em] = 169\frac{1}{2}.\end{aligned}$$

Hence, 3(x² + y²) = $169\frac{1}{2}.$

(iii) From parts (i) and (ii) we get,

(x - y) = ±7 and x² + y² = $\frac{113}{2}$.

When (x - y) = 7,

Substituting values we get,

$$\begin{aligned}5(x^2 + y^2) + 4(x - y) = 5 \times \frac{113}{2} + 4 \times 7 \\[1em] = \frac{565}{2} + 28 \\[1em] = \frac{565 + 56}{2} \\[1em] = \frac{621}{2} \\[1em] = 310\frac{1}{2}.\end{aligned}$$

When (x - y) = -7,

Substituting values we get,

$$\begin{aligned}5(x^2 + y^2) + 4(x - y) = 5 \times \frac{113}{2} + 4 \times -7 \\[1em] = \frac{565}{2} - 28 \\[1em] = \frac{565 - 56}{2} \\[1em] = \frac{509}{2} \\[1em] = 254\frac{1}{2}.\end{aligned}$$

Hence, 5(x² + y²) + 4(x - y) = $310\frac{1}{2} \text{ or } 254\frac{1}{2}$.

We know that,

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

Substituting values in above equation,

$$\begin{aligned}= 3 \times 34 + 2 \times 10\frac{1}{2} \\[1em] = 102 + 2 \times \frac{21}{2} \\[1em] = 102 + 21 \\[1em] = 123.\end{aligned}$$

Hence, 2(x + y)² + (x - y)² = 123.

We know that,

⇒ (a - b)³ = a³ - b³ - 3ab(a - b).

∴ a³ - b³ = (a - b)³ + 3ab(a - b).

Substituting values we get,

⇒ a³ - b³ = (3)³ + 3 × 4 × 3
⇒ a³ - b³ = 27 + 36 = 63.

Hence, a³ - b³ = 63.

We know that,

⇒ (a - b)³ = a³ - b³ - 3ab(a - b).

∴ a³ - b³ = (a - b)³ + 3ab(a - b).

∴ 8a³ - 27b³ = (2a)³ - (3b)³ = (2a - 3b)³ + 3 × 2a × 3b (2a - 3b)

Substituting values we get,

⇒ 8a³ - 27b³ = (2a - 3b)³ + 3 × 2a × 3b (2a - 3b)

⇒ 8a³ - 27b³ = 3³ + 18ab × 3

⇒ 8a³ - 27b³ = 27 + 18 × 2 × 3

⇒ 8a³ - 27b³ = 27 + 108

⇒ 8a³ - 27b³ = 135.

Hence, 8a³ - 27b³ = 135.

(i) We know that,

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

Substituting values we get,

$$\begin{aligned}x^2 + \frac{1}{x^2} = 4^2 - 2 \\[1em] = 16 - 2 \\[1em] = 14.\end{aligned}$$

Hence, $x^2 + \frac{1}{x^2}$ = 14.

(ii) We know that,

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

Substituting values we get,

$$\begin{aligned}x^4 + \frac{1}{x^4} = 14^2 - 2 \\[1em] = 196 - 2 \\[1em] = 194.\end{aligned}$$

Hence, $x^4 + \frac{1}{x^4}$ = 194.

(iii) We know that,

$x^3 + \frac{1}{x^3} = \Big(x + \frac{1}{x}\Big)^3 - 3 \times x \times \frac{1}{x}\Big(x + \frac{1}{x}\Big)$

Substituting values we get,

$$\begin{aligned}x^3 + \frac{1}{x^3} = (4)^3 - 3 \times 4 \\[1em] = 64 - 12 \\[1em] = 52.\end{aligned}$$

Hence, $x^3 + \frac{1}{x^3}$ = 52.

(iv) We know that,

$$\begin{aligned}\Rightarrow \Big(x - \frac{1}{x}\Big)^2 = x^2 + \frac{1}{x^2} - 2 \\[1em] \therefore x - \frac{1}{x} = \sqrt{x^2 + \frac{1}{x^2} - 2}\end{aligned}$$

Substituting values we get,

$$\begin{aligned}x - \frac{1}{x} = \sqrt{14 - 2} \\[1em] = \sqrt{12} \\[1em] = \sqrt{4 \times 3} \\[1em] = \pm 2\sqrt{3}.\end{aligned}$$

Hence, $x - \frac{1}{x} = \pm 2\sqrt{3}$.

$$\begin{aligned}x^4 + \frac{1}{x^4} = (x^2)^2 + \Big(\frac{1}{x^2}\Big)^2 .....[i] \\[1em] \Big(x^2 + \frac{1}{x^2}\Big)^2 = (x^2)^2 + 2 \times (x^2)^2 \times \frac{1}{x^2} + \Big(\frac{1}{x^2}\Big)^2 \\[1em] \therefore (x^2)^2 + \Big(\frac{1}{x^2}\Big)^2 = \Big(x^2 + \frac{1}{x^2}\Big)^2 - 2 \\[1em]\end{aligned}$$

Putting this value of $(x^2)^2 + \Big(\frac{1}{x^2}\Big)^2$ in eqn (i), we get:

$$\begin{aligned}x^4 + \frac{1}{x^4} = \Big(x^2 + \frac{1}{x^2}\Big)^2 - 2 \space .....[ii] \\[1em] \Big(x - \frac{1}{x}\Big)^2 = x^2 - 2 \times x \times \frac{1}{x} + \Big(\frac{1}{x}\Big)^2 \\[1em] \therefore x^2 + \frac{1}{x^2} = \Big(x - \frac{1}{x}\Big)^2 + 2 \\[1em]\end{aligned}$$

Putting this value of $x^2 + \frac{1}{x^2}$ in eqn (ii), we get:

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

Putting the given values in above eqn, we get:

$$\begin{aligned}x^4 + \frac{1}{x^4} = (5^2 + 2)^2 - 2 \\[1em] = (27)^2 - 2 \\[1em] = 729 - 2 \\[1em] = 727.\end{aligned}$$

Hence, $x^4 + \frac{1}{x^4}$ = 727.

(i) We know that,

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

Substituting values we get,

$$\begin{aligned}x^2 + \frac{1}{x^2} = (\sqrt{5})^2 + 2 \\[1em] = 5 + 2 \\[1em] = 7.\end{aligned}$$

Hence, $x^2 + \frac{1}{x^2}$ = 7.

(ii) We know that,

$$\begin{aligned}\Rightarrow \Big(x + \frac{1}{x}\Big)^2 = x^2 + \frac{1}{x^2} + 2 \\[1em] \therefore x + \frac{1}{x} = \sqrt{x^2 + \frac{1}{x^2} + 2}\end{aligned}$$

Substituting values we get,

$$\begin{aligned}x + \frac{1}{x} = \sqrt{7 + 2} \\[1em] = \sqrt{9} \\[1em] = \pm 3.\end{aligned}$$

Hence, $x + \frac{1}{x} = \pm 3.$

(iii) We know that,

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

When $x + \frac{1}{x} = 3$.

Substituting values we get,

$$\begin{aligned}x^3 + \frac{1}{x^3} = 3^3 - 3 \times 3 \\[1em] = 27 - 9 \\[1em] = 18.\end{aligned}$$

When $x + \frac{1}{x} = -3$.

Substituting values we get,

$$\begin{aligned}x^3 + \frac{1}{x^3} = (-3)^3 - 3 \times (-3) \\[1em] = -27 + 9 \\[1em] = -18.\end{aligned}$$

Hence, $x^3 + \frac{1}{x^3} = \pm 18$.

(i) We know that,

$$\begin{aligned}\Rightarrow \Big(x - \frac{1}{x}\Big)^2 = x^2 + \frac{1}{x^2} - 2 \text{ ......(i)} \\[1em] \Rightarrow \Big(x + \frac{1}{x}\Big)^2 = x^2 + \frac{1}{x^2} + 2 \text{ ......(ii)}\end{aligned}$$

Eq. (i) can be written as,

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

Substituting values we get,

$$\begin{aligned}x - \frac{1}{x} = \sqrt{(6)^2 - 4} \\[1em] = \sqrt{36 - 4} \\[1em] = \sqrt{32} \\[1em] = \pm 4\sqrt{2}.\end{aligned}$$

Hence, $x - \frac{1}{x} = \pm 4\sqrt{2}.$

(ii) We know that,

$x^2 - \frac{1}{x^2} = \Big(x + \frac{1}{x}\Big)\Big(x - \frac{1}{x}\Big).$

When, $x - \frac{1}{x} = 4\sqrt{2}$

Substituting values we get,

$x^2 - \frac{1}{x^2} = 6 \times 4\sqrt{2} = 24\sqrt{2}.$

When, $x - \frac{1}{x} = -4\sqrt{2}$

Substituting values we get,

$x^2 - \frac{1}{x^2} = 6 \times -4\sqrt{2} = -24\sqrt{2}.$

Hence, $x^2 - \frac{1}{x^2} = ±24\sqrt{2}.$

We know that,

$x^2 + \frac{1}{x^2} = \Big(x + \frac{1}{x}\Big)^2 - 2$.

Substituting values we get,

$x^2 + \frac{1}{x^2} = 2^2 - 2 = 4 - 2 = 2$ ........(i)

We know that,

$x^3 + \frac{1}{x^3} = \Big(x + \frac{1}{x}\Big)^3 - 3\Big(x + \frac{1}{x}\Big)$

Substituting values we get,

$x^3 + \frac{1}{x^3} = (2)^3 - 3 \times 2 = 8 - 6 = 2$ .......(ii)

We know that,

$x^4 + \frac{1}{x^4} = \Big(x^2 + \frac{1}{x^2}\Big)^2 - 2$

Substituting values we get,

$x^4 + \frac{1}{x^4} = 2^2 - 2 = 4 - 2 = 2$ ...........(iii)

From (i), (ii) and (iii),

Hence proved, $x^2 + \frac{1}{x^2} = x^3 + \frac{1}{x^3} = x^4 + \frac{1}{x^4}.$ when $x + \frac{1}{x} = 2$

We know that,

⇒ (a - b)³ = a³ - b³ - 3ab(a - b)

⇒ a³ - b³ = (a - b)³ + 3ab(a - b).

$$\begin{aligned}x^3 - \frac{8}{x^3} = (x)^3 - \Big(\frac{2}{x}\Big)^3 \\[1em] = \Big(x - \frac{2}{x}\Big)^3 + 3 \times x \times \frac{2}{x}\Big(x - \frac{2}{x}\Big) \\[1em] = \Big(x - \frac{2}{x}\Big)^3 + 6\Big(x - \frac{2}{x}\Big).\end{aligned}$$

Substituting values we get,

$x^3 - \frac{8}{x^3} = 3^3 + 6 \times 3 = 27 + 18 = 45.$

Hence, the value of $x^3 - \frac{8}{x^3} = 45.$

We know that,

⇒ (a + 2b)³ = a³ + 8b³ + 3(a)(2b)(a + 2b)

Substituting values in above formula,

⇒ 5³ = a³ + 8b³ + 6ab × 5
⇒ 125 = a³ + 8b³ + 30ab.

Hence, proved that a³ + 8b³ + 30ab = 125.

We know that,

$$\begin{aligned}\Rightarrow \Big(a + \frac{1}{a}\Big)^3 = a^3 + \frac{1}{a^3} + 3\Big(a + \frac{1}{a}\Big) \\[1em] \Rightarrow a^3 + \frac{1}{a^3} = \Big(a + \frac{1}{a}\Big)^3 - 3\Big(a + \frac{1}{a}\Big).\end{aligned}$$

Substituting values we get,

$$\begin{aligned}a^3 + \frac{1}{a^3} = p^3 - 3p \\[1em] = p(p^2 - 3).\end{aligned}$$

Hence, proved that $a^3 + \frac{1}{a^3} = p(p^2 - 3)$.

We know that,

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

Substituting values we get,

$x - \frac{1}{x} = \sqrt{27 - 2} = \sqrt{25} = \pm 5.$

Here, $3x^3 + 5x - \frac{3}{x^3} - \frac{5}{x}$ can be written as,

$$\begin{aligned}3x^3 + 5x - \frac{3}{x^3} - \frac{5}{x} = 3\Big(x^3 - \frac{1}{x^3}\Big) + 5\Big(x - \frac{1}{x}\Big) \\[1em] = 3\Big[\Big(x - \frac{1}{x}\Big)^3 + 3\Big(x - \frac{1}{x}\Big)\Big] + 5\Big(x - \frac{1}{x}\Big).\end{aligned}$$

Considering $x - \frac{1}{x} = 5$ in first case and substituting value we get,

$$\begin{aligned}3x^3 + 5x - \frac{3}{x^3} - \frac{5}{x} = 3(5^3 + 3 \times 5) + (5 \times 5) \\[1em] = 3(125 + 15) + 25 \\[1em] = (3 \times 140) + 25 \\[1em] = 420 + 25 \\[1em] = 445.\end{aligned}$$

Considering $x - \frac{1}{x} = -5$ in second case and substituting value we get,

$$\begin{aligned}3x^3 + 5x - \frac{3}{x^3} - \frac{5}{x} = 3[(-5)^3 + 3 \times (-5)] + [5 \times (-5)] \\[1em] = 3(-125 - 15) - 25 \\[1em] = (3 \times -140) - 25 \\[1em] = -420 - 25 \\[1em] = -445.\end{aligned}$$

Hence, $3x^3 + 5x - \frac{3}{x^3} - \frac{5}{x} = \pm 445.$

$$\begin{aligned}x^2 + \frac{1}{25x^2} = (x)^2 + \frac{1}{(5x)^2} ....[i] \\[1em] \Big(x + \frac{1}{5x}\Big)^2 = (x)^2 + 2 \times x \times \frac{1}{5x} + \frac{1}{(5x)^2} \\[1em] = (x)^2 + \frac{1}{(5x)^2} + \frac{2}{5} \\[1em] \therefore x^2 + \frac{1}{25x^2} = \Big(x + \frac{1}{5x}\Big)^2 - \frac{2}{5}\end{aligned}$$

Putting this value of $x^2 + \frac{1}{25x^2}$ in eqn (i), we get:

$x^2 + \frac{1}{25x^2} = \Big(x + \frac{1}{5x}\Big)^2 - \frac{2}{5}$

Let $x + \frac{1}{5x}$ be a.

Substituting values we get,

$$\begin{aligned}\Rightarrow 8\frac{3}{5} = a^2 - \frac{2}{5} \\[1em] \Rightarrow \frac{43}{5} = \frac{5a^2 - 2}{5} \\[1em] \Rightarrow 5a^2 - 2 = 43 \\[1em] \Rightarrow 5a^2 = 45 \\[1em] \Rightarrow a^2 = 9 \\[1em] \Rightarrow a = \pm 3.\end{aligned}$$

Hence, $x + \frac{1}{5x} = \pm 3$.

We know that,

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

Substituting values in above equation we get,

$$\begin{aligned}\Rightarrow 8 = \Big(x + \frac{1}{2x}\Big)^2 - 1 \\[1em] \Rightarrow \Big(x + \frac{1}{2x}\Big)^2 = 8 + 1 \\[1em] \Rightarrow \Big(x + \frac{1}{2x}\Big)^2 = 9 \\[1em] \Rightarrow x + \frac{1}{2x} = \sqrt{9} \\[1em] \Rightarrow x + \frac{1}{2x} = \pm 3.\end{aligned}$$

We know that,

$$\begin{aligned}\Rightarrow (a + b)^3 = a^3 + b^3 + 3ab(a + b) \\[1em] \Rightarrow \Big(x + \frac{1}{2x}\Big)^3 = x^3 + \Big(\frac{1}{2x}\Big)^3 + 3 \times x \times \frac{1}{2x}\Big(x + \frac{1}{2x}\Big) \\[1em] \Rightarrow \Big(x + \frac{1}{2x}\Big)^3 = x^3 + \frac{1}{8x^3} + \frac{3}{2}\Big(x + \frac{1}{2x}\Big) \\[1em] \therefore x^3 + \frac{1}{8x^3} = \Big(x + \frac{1}{2x}\Big)^3 - \frac{3}{2}\Big(x + \frac{1}{2x}\Big).\end{aligned}$$.

Case 1: $x + \frac{1}{2x} = 3$ substituting values we get,

$$\begin{aligned}x^3 + \frac{1}{8x^3} = 3^3 - \frac{3}{2} \times 3 \\[1em] = 27 - \frac{9}{2} \\[1em] = \frac{54 - 9}{2} \\[1em] = \frac{45}{2}.\end{aligned}$$

Case 2 : $x + \frac{1}{2x} = -3$, substituting values we get,

$$\begin{aligned}x^3 + \frac{1}{8x^3} = (-3)^3 - \frac{3}{2} \times (-3) \\[1em] = -27 + \frac{9}{2} \\[1em] = \frac{-54 + 9}{2} \\[1em] = -\frac{45}{2}.\end{aligned}$$

Hence, $x^3 + \frac{1}{8x^3} = \pm \frac{45}{2} = \pm 22\frac{1}{2}$.

Dividing each term of a² - 3a + 1 = 0 by a we get,

$a + \frac{1}{a} = 3$.

(i) We know that,

$a^2 + \frac{1}{a^2} = \Big(a + \frac{1}{a}\Big)^2 - 2$.

Substituting values we get,

$a^2 + \frac{1}{a^2} = 3^2 - 2 = 9 - 2 = 7.$

Hence, $a^2 + \frac{1}{a^2}$ = 7.

(ii) We know that,

$a^3 + \frac{1}{a^3} = \Big(a + \frac{1}{a}\Big)^3 - 3\Big(a + \frac{1}{a}\Big)$.

Substituting values we get,

$a^3 + \frac{1}{a^3} = 3^3 - 3 \times 3 = 27 - 9 = 18.$

Hence, $a^3 + \frac{1}{a^3} = 18.$

Given,

$$\begin{aligned}\Rightarrow a = \frac{1}{a - 5} \\[1em] \therefore a(a - 5) = 1 \\[1em] \Rightarrow a^2 - 5a = 1 \\[1em] \Rightarrow a^2 - 5a - 1 = 0\end{aligned}$$

Now,

(i) Dividing above equation by a we get,

$$\begin{aligned}\Rightarrow a - 5 - \frac{1}{a} = 0 \\[1em] \Rightarrow a - \frac{1}{a} = 5.\end{aligned}$$.

Hence, $a - \frac{1}{a} = 5$.

(ii) We know that,

$$\begin{aligned}\Rightarrow \Big(a + \frac{1}{a}\Big)^2 = a^2 + \frac{1}{a^2} + 2 \space .....(i) \\[1em] \Rightarrow \Big(a - \frac{1}{a}\Big)^2 = a^2 + \frac{1}{a^2} - 2 \space .....(ii) \\[1em]\end{aligned}$$

Subtracting eq. (ii) from (i) we get,

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

Substituting values we get,

$$\begin{aligned}a + \frac{1}{a} = \sqrt{5^2 + 4} \\[1em] = \sqrt{25 + 4} \\[1em] = \pm \sqrt{29}.\end{aligned}$$

Hence, $a + \frac{1}{a} = \pm \sqrt{29}$.

(iii) We know that,

$\Big(a^2 - \frac{1}{a^2}\Big) = \Big(a + \frac{1}{a}\Big)\Big(a - \frac{1}{a}\Big)$

Substituting values we get,

$\Big(a^2 - \frac{1}{a^2}\Big) = \pm \sqrt{29} \times 5 = \pm 5\sqrt{29}.$

Hence, $a^2 - \frac{1}{a^2} = \pm 5\sqrt{29}$.

Given,

$$\begin{aligned}\Rightarrow \Big(x + \frac{1}{x}\Big)^2 = 3 \\[1em] \therefore x + \frac{1}{x} = \pm \sqrt{3}.\end{aligned}$$

We know that,

$x^3 + \frac{1}{x^3} = \Big(x + \frac{1}{x}\Big)^3 - 3\Big(x + \frac{1}{x}\Big)$

When $x + \frac{1}{x} = \sqrt 3$ substituting values we get,

$x^3 + \frac{1}{x^3} = (\sqrt{3})^3 - 3 \times \sqrt{3} = 3\sqrt{3} - 3\sqrt{3} = 0.$

When $x + \frac{1}{x} = -\sqrt{3}$ substituting values we get,

$x^3 + \frac{1}{x^3} = (-\sqrt{3})^3 - 3 \times (-\sqrt{3}) = -3\sqrt{3} + 3\sqrt{3} = 0.$

Hence, $x^3 + \frac{1}{x^3} = 0$.

Given,

$$\begin{aligned}x = 5 - 2\sqrt{6} \\[1em] \Rightarrow \frac{1}{x} = \frac{1}{5- 2\sqrt{6}} \\[1em] \Rightarrow \frac{1}{x} = \frac{1}{5 - 2\sqrt{6}} \times \frac{5 + 2\sqrt{6}}{5 + 2\sqrt{6}} \\[1em] \Rightarrow \frac{1}{x} = \frac{5 + 2\sqrt{6}}{(5 - 2\sqrt{6})(5 + 2\sqrt{6})} \\[1em] \Rightarrow \frac{1}{x} = \frac{5 + 2\sqrt{6}}{5^2 - (2\sqrt{6})^2} \\[1em] \Rightarrow \frac{1}{x} = \frac{5 + 2\sqrt{6}}{25 - 24} \\[1em] \Rightarrow \frac{1}{x} = \frac{5 + 2\sqrt{6}}{1} = 5 + 2\sqrt{6}.\end{aligned}$$

So,

$x + \frac{1}{x} = 5 - 2\sqrt{6} + 5 + 2\sqrt{6} = 10.$

We know that,

$$\begin{aligned}\Rightarrow \Big(\sqrt{x} + \frac{1}{\sqrt{x}}\Big)^2 = x + \frac{1}{x} + 2 \times \sqrt{x} \times \frac{1}{\sqrt{x}} \\[1em] \Rightarrow \Big(\sqrt{x} + \frac{1}{\sqrt{x}}\Big)^2 = x + \frac{1}{x} + 2 \\[1em] \Rightarrow \sqrt{x} + \frac{1}{\sqrt{x}} = \sqrt{x + \frac{1}{x} + 2} = \sqrt{10 + 2} = \sqrt{12} = \pm 2\sqrt{3}.\end{aligned}$$

Hence, $\sqrt{x} + \frac{1}{\sqrt{x}} = \pm 2\sqrt{3}.$

We know that,

a² + b² + c² = (a + b + c)² - 2(ab + bc + ca)

Substituting values we get,

a² + b² + c² = (12)² - 2(22) = 144 - 44 = 100.

Hence, a² + b² + c² = 100.

We know that,

a² + b² + c² = (a + b + c)² - 2(ab + bc + ca)

Substituting values we get,

⇒ 100 = 12² - 2(ab + bc + ca)

⇒ 100 = 144 - 2(ab + bc + ca)

⇒ 2(ab + bc + ca) = 144 - 100

⇒ 2(ab + bc + ca) = 44

⇒ (ab + bc + ca) = 22.

Hence, ab + bc + ca = 22.

We know that,

(a + b + c)² = a² + b² + c² + 2(ab + bc + ca)

Substituting values we get,

⇒ (a + b + c)² = 125 + 2(50) = 125 + 100 = 225.

$\therefore (a + b + c) = \sqrt{225} = \pm 15.$

Hence, (a + b + c) = ± 15.

Given, a + b - c = 5. Squaring both sides we get,

⇒ (a + b - c)² = 5²

⇒ a² + b² + c² + 2ab - 2bc - 2ca = 25

Substituting values we get,

⇒ 29 + 2(ab - bc - ca) = 25

⇒ 2(ab - bc - ca) = 25 - 29

⇒ 2(ab - bc - ca) = -4

⇒ (ab - bc - ca) = -2.

Hence, (ab - bc - ca) = -2.

We know that,

⇒ (a - b)² = a² + b² - 2ab

⇒ (7)² = 85 - 2ab

⇒ 49 = 85 - 2ab

⇒ 2ab = 85 - 49

⇒ 2ab = 36

⇒ ab = 18.

We know that,

a³ - b³ = (a - b)(a² + b² + ab)

Substituting values we get,

⇒ a³ - b³ = 7(85 + 18) = 7(103) = 721.

Hence, a³ - b³ = 721.

Given,

x = y - 3 or x - y = -3 and

x² + y² = 29.

We know that,

⇒ (x + y)² = x² + y² + 2xy ......(i)

⇒ (x - y)² = x² + y² - 2xy .......(ii)

Subtracting eq. (ii) from (i) we get,

⇒ (x + y)² - (x - y)² = x² + y² + 2xy - (x² + y² - 2xy)

⇒ (x + y)² - (x - y)² = 4xy

⇒ x² + y² = (x - y)² + 4xy

Substituting values we get,

⇒ 29 = (-3)² + 4xy
⇒ 29 = 9 + 4xy
⇒ 29 - 9 = 4xy
⇒ 4xy = 20
⇒ xy = 5.

Hence, xy = 5.

Let two numbers be x and y.

Given,

x + y = 8 and xy = 15.

We know that,

⇒ x³ + y³ = (x + y)³ - 3xy(x + y)

⇒ x³ + y³ = 8³ - 3(15)(8)

⇒ x³ + y³ = 512 - 360

⇒ x³ + y³ = 152.

Hence, x³ + y³ = 152.

We know that,

$x^2 + \frac{1}{x^2} = \Big(x + \frac{1}{x}\Big)^2 -2$.

Substituting values we get,

$x^2 + \frac{1}{x^2} = (2)^2 - 2 = 4 - 2 = 2$.

Hence, Option 2 is the correct option.

We know that,

$(x + y) = \sqrt{x^2 + y^2 + 2xy}$.

Substituting values we get,

$(x + y) = \sqrt{9 + 2 \times 8} = \sqrt{9 + 16} = \sqrt{25} = \pm 5$.

Hence, Option 4 is the correct option.

Applying a² - b² = (a - b)(a + b) formula to (102)² - (98)² we get,

(102)² - (98)² = (102 - 98)(102 + 98) = 4 × 200 = 800.

Hence, Option 4 is the correct option.

We can write 96 × 104 as (100 - 4) × (100 + 4).

By a² - b² = (a - b)(a + b) formula,

(100 - 4) × (100 + 4) = (100)² - (4)² = 10000 - 16 = 9984.

Hence, Option 1 is the correct option.

By a² - b² = (a - b)(a + b) formula,

$$\begin{aligned}\frac{103^2 - 97^2}{200} = \frac{(103 - 97)(103 + 97)}{200} \\[1em] = \frac{6 \times 200}{200} \\[1em] = 6.\end{aligned}$$

Hence, Option 4 is the correct option.

We know that,

x² + y² = (x + y)² - 2xy.

Substituting values we get,

x² + y² = (11)² - 2 × 24 = 121 - 48 = 73.

Hence, Option 2 is the correct option.

By a² - b² = (a - b)(a + b) formula,

249² - 248² = (249 - 248)(249 + 248) = 1 × 497 = 497.

Hence, Option 4 is the correct option.

Given,

$$\begin{aligned}\Rightarrow \frac{x}{y} + \frac{y}{x} = -1 \\[1em] \Rightarrow \frac{x^2 + y^2}{xy} = -1 \\[1em] \Rightarrow x^2 + y^2 = -xy \\[1em] \Rightarrow x^2 + y^2 + xy = 0.\end{aligned}$$

We know that,

x³ - y³ = (x - y)(x² + y² + xy) = (x - y) × 0 = 0.

Hence, Option 3 is the correct option.

We know that,

a³ + b³ + c³ - 3abc = (a + b + c)(a² + b² + c² - bc - ca - ab).

Substituting values we get,

a³ + b³ + c³ - 3abc = 0 × (a² + b² + c² - bc - ca - ab)

a³ + b³ + c³ - 3abc = 0

a³ + b³ + c³ = 3abc.

Hence, Option 4 is the correct option.

We know that,

(a - b)³ = a³ -3ab(a - b) - b³

⇒ a³ - b³ = (a - b)³ + 3ab(a - b)

$\therefore x^3 - \frac{8}{x^3} = \Big(x - \frac{2}{x}\Big)^3 + 6\Big(x - \frac{2}{x}\Big)$

Substituting values we get,

$$\begin{aligned}x^3 - \frac{8}{x^3} = 3^3 + 6 \times 3 \\[1em] = 27 + 18 \\[1em] = 45.\end{aligned}$$

Hence, Option 3 is the correct option.

Given,

⇒ a + b = 0

Squaring both sides, we get

⇒ (a + b)² = 0²

⇒ a² + b² + 2ab = 0

⇒ a² + b² = -2ab

∴ Statement 1 is false.

(a + b)² = a² + b² + 2ab

∴ Statement 2 is false.

∴ Both statements are false.

Hence, option 2 is the correct option.

Given,

⇒ x = $\frac{1}{x}$

⇒ x.x = 1

⇒ x² = 1

⇒ x = $\sqrt{1}$

⇒ x = ± 1

∴ Assertion (A) is true.

Solving,

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

∴ Reason (R) is true.

∴ Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct reason (or explanation) for Assertion (A).

Hence, option 4 is the correct option.

Given,

⇒ (a - b)(a + b)

⇒ a(a + b) - b(a + b)

⇒ a² + ab - ab - b²

⇒ a² - b²

∴ Reason (R) is true.

Solving,

⇒ 1003 x 997

⇒ (1000 + 3)(1000 - 3)

⇒ 1000² - 3²

⇒ 1000000 - 9

⇒ 999991.

∴ Assertion (A) is true.

∴ Both Assertion (A) and Reason (R) are true, and Reason (R) is the correct reason for Assertion (A).

Hence, option 3 is the correct option.

(i) On solving,

⇒ (2x + 3y + 5)(2x + 3y - 5) = (2x + 3y)² - 5²

⇒ (2x + 3y + 5)(2x + 3y - 5) = 4x² + 9y² + 12xy - 25.

Hence, (2x + 3y + 5)(2x + 3y - 5) = 4x² + 9y² + 12xy - 25.

(ii) On solving,

⇒ (6 - 4a - 7b)² = (6 - 4a)² + (7b)² - 2(6 - 4a)(7b)

⇒ (6 - 4a - 7b)² = 36 + 16a² - 48a + 49b² - 14b(6 - 4a)

⇒ (6 - 4a - 7b)² = 36 + 16a² + 49b² - 48a + 56ab - 84b.

Hence, (6 - 4a - 7b)² = 36 + 16a² + 49b² - 48a + 56ab - 84b.

(iii) On solving,

⇒ (7 - 3xy)³ = (7)³ - (3xy)³ - 3(7)(3xy)(7 - 3xy)

⇒ (7 - 3xy)³ = 343 - 27x³y³ - 63xy(7 - 3xy)

⇒ (7 - 3xy)³ = 343 - 27x³y³ - 441xy + 189x²y².

Hence, (7 - 3xy)³ = 343 - 27x³y³ - 441xy + 189x²y².

(iv) On solving,

⇒ (x + y + 2)³ = (x)³ + (y + 2)³ + 3(x)(y + 2)(x + y + 2)

⇒ (x + y + 2)³ = x³ + y³ + 2³ + 3(y)(2)(y + 2) + (3xy + 6x)(x + y + 2)

⇒ (x + y + 2)³ = x³ + y³ + 8 + 6y(y + 2) + 3xy(x + y + 2) + 6x(x + y + 2)

⇒ (x + y + 2)³ = x³ + y³ + 8 + 6y² + 12y + 3x²y + 3xy² + 6xy + 6x² + 6xy + 12x

⇒ (x + y + 2)³ = x³ + y³ + 3x²y + 3xy² + 6x² + 6y² + 12xy + 12x + 12y + 8.

Hence, (x + y + 2)³ = x³ + y³ + 3x²y + 3xy² + 6x² + 6y² + 12xy + 12x + 12y + 8.

As (a - b)(a + b) = (a² - b²)

On solving,

$$\begin{aligned}(x - 2)(x + 2)(x^2 + 4)(x^2 + 16) = (x^2 - 4)(x^2 + 4)(x^4 + 16) \\[1em] = (x^4 - 16)(x^4 + 16) \\[1em] = ((x^4)^2 - 16^2) \\[1em] = x^8 - 256.\end{aligned}$$

On simplifying we get, (x - 2)(x + 2)(x² + 4)(x⁴ + 16) = x⁸ - 256.

We can write 1002 × 998 as (1000 + 2)(1000 - 2).

As (a - b)(a + b) = (a² - b²)

We get,

(1000 + 2)(1000 - 2) = 1000² - 2² = 1000000 - 4 = 999996.

Hence, 1002 × 998 = 999996.

Given,

a + 2b + 3c = 0
a + 2b = -3c.

Cubing both sides,

(a + 2b)³ = (-3c)³

a³ + (2b)³ + 3(a)(2b)(a + 2b) = -27c³

a³ + 8b³ + 6ab(-3c) = -27c³

a³ + 8b³ - 18abc = -27c³

a³ + 8b³ + 27c³ = 18abc.

Hence, proved that a³ + 8b³ + 27c³ = 18abc.

Given, 2x = 3y - 5 or 2x - 3y = -5

Cubing both sides we get,

⇒ (2x - 3y)³ = (-5)³

(2x)³ - (3y)³ - 3(2x)(3y)(2x - 3y) = -125

⇒ 8x³ - 27y³ - 18xy(2x - 3y) = -125

⇒ 8x³ - 27y³ - 18xy(-5) = -125

⇒ 8x³ - 27y³ + 90xy = -125

⇒ 8x³ - 27y³ + 90xy + 125 = 0.

Hence, 8x³ - 27y³ + 90xy + 125 = 0.

We know that,

$a^4 + \frac{1}{a^4} = \Big(a^2 - \frac{1}{a^2}\Big)^2 + 2$

Substituting values we get,

$a^4 + \frac{1}{a^4} = (5)^2 + 2 = 25 + 2 = 27.$

Hence, $a^4 + \frac{1}{a^4} = 27.$

We know that,

$$\begin{aligned}\Rightarrow \Big(a + \frac{1}{a}\Big)^2 = a^2 + \frac{1}{a^2} + 2 \space .....(i)\\[1em] \Rightarrow \Big(a - \frac{1}{a}\Big)^2 = a^2 + \frac{1}{a^2} - 2 \space .....(ii)\end{aligned}$$

Subtracting eq. (ii) from (i) we get,

$$\begin{aligned}\Rightarrow \Big(a + \frac{1}{a}\Big)^2 - \Big(a - \frac{1}{a}\Big)^2 = 4 \\[1em] \text{Substituting values we get}, \\[1em] \Rightarrow p^2 - q^2 = 4.\end{aligned}$$

Hence, p² - q² = 4.

Given,

$$\begin{aligned}\phantom{\therefore} \frac{a^2 + 1}{a} = 4 \\[1em] \therefore a + \frac{1}{a} = 4.\end{aligned}$$

Solving,

$$\begin{aligned}\Rightarrow 2a^3 + \frac{2}{a^3} = 2\Big(a^3 + \frac{1}{a^3}\Big) \\[1em] = 2\Big[\Big(a + \frac{1}{a}\Big)^3 - 3\Big(a + \frac{1}{a}\Big)\Big] \\[1em] = 2[4^3 - 3 \times 4] \\[1em] = 2[64 - 12] \\[1em] = 2 \times 52 \\[1em] = 104.\end{aligned}$$

Hence, the value of $2a^3 + \frac{2}{a^3}$ = 104.

(i) Given,

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

On dividing above equation by x,

$$\begin{aligned}\Rightarrow 4 - x = \frac{1}{x} \\[1em] \Rightarrow x + \frac{1}{x} = 4.\end{aligned}$$

Hence, the value of $x + \frac{1}{x}$ = 4.

(ii) We know that,

$\Big(x^3 + \frac{1}{x^3}\Big) = \Big(x + \frac{1}{x}\Big)^3 - 3\Big(x + \frac{1}{x}\Big)$.

Substituting values we get,

$\Big(x^3 + \frac{1}{x^3}\Big) = 4^3 - 3 \times 4 = 64 - 12 = 52.$

Hence, the value of $x^3 + \frac{1}{x^3}$ = 52.

(iii) We know,

$\Big(x^6 + \frac{1}{x^6}\Big) = \Big(x^3 + \frac{1}{x^3}\Big)^2 - 2$.

Substituting values we get,

$\Big(x^6 + \frac{1}{x^6}\Big) = 52^2 - 2 = 2704 - 2 = 2702.$

Hence, the value of $x^6 + \frac{1}{x^6}$ = 2702.

We know that,

$x^3 - \frac{1}{x^3} = \Big(x - \frac{1}{x}\Big)^3 + 3\Big(x - \frac{1}{x}\Big)$

Substituting values we get,

$$\begin{aligned}x^3 - \frac{1}{x^3} = (3 + 2\sqrt{2})^3 + 3(3 + 2\sqrt{2}) \\[1em] = (3)^3 + (2\sqrt{2})^3 + 3(3)(2\sqrt{2})(3 + 2\sqrt{2}) + 9 + 6\sqrt{2} \\[1em] = 27 + 16\sqrt{2} + 18\sqrt{2}(3 + 2\sqrt{2}) + 9 + 6\sqrt{2} \\[1em] = 27 + 9 + 16\sqrt{2} + 54\sqrt{2} + 72 + 6\sqrt{2} \\[1em] = 108 + 76\sqrt{2}.\end{aligned}$$

So, $\frac{1}{4}\Big(x^3 - \frac{1}{x^3}\Big) = \frac{1}{4} \times (108 + 76\sqrt{2}) = 27 + 19\sqrt{2}$.

Hence, $\frac{1}{4}\Big(x^3 - \frac{1}{x^3}\Big) = 27 + 19\sqrt{2}$.

We know that,

$x - \frac{1}{x} = \sqrt{\Big(x + \frac{1}{x}\Big)^2 - 4}$

Substituting values we get,

$$\begin{aligned}x - \frac{1}{x} = \sqrt{\Big(3\frac{1}{3}\Big)^2 - 4} \\[1em] = \sqrt{\Big(\frac{10}{3}\Big)^2 - 4} \\[1em] = \sqrt{\frac{100}{9} - 4} \\[1em] = \sqrt{\frac{100 - 36}{9}} \\[1em] = \sqrt{\frac{64}{9}} \\[1em] = \pm \frac{8}{3}.\end{aligned}$$

We know that,

$x^3 - \frac{1}{x^3} = \Big(x - \frac{1}{x}\Big)^3 + 3\Big(x - \frac{1}{x}\Big)$.

Substituting values we get,

$$\begin{aligned}x^3 - \frac{1}{x^3} = \Big(\pm \frac{8}{3}\Big)^3 + 3 \times \pm \Big(\frac{8}{3}\Big) \\[1em] = \pm \frac{512}{27} \pm 8 \\[1em] = \frac{\pm 512 \pm 216}{27} \\[1em] = \pm \frac{728}{27} \\[1em] = \pm 26\frac{26}{27}.\end{aligned}$$

Hence, the value of $x^3 - \frac{1}{x^3} = \pm 26\frac{26}{27}$.

Given,

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

So,

$x - \frac{1}{x} = 2 - \sqrt{3} - (2 + \sqrt{3}) = -2\sqrt{3}.$

Cubing both sides we get,

$$\begin{aligned}\Rightarrow \Big(x - \frac{1}{x}\Big)^3 = (-2\sqrt{3})^3 \\[1em] \Rightarrow x^3 - \frac{1}{x^3} - 3(x)\Big(\frac{1}{x}\Big)\Big(x - \frac{1}{x}\Big) = -24\sqrt{3} \\[1em] \Rightarrow x^3 - \frac{1}{x^3} - 3 \times -2\sqrt{3} = -24\sqrt{3} \\[1em] \Rightarrow x^3 - \frac{1}{x^3} + 6\sqrt{3} = -24\sqrt{3} \\[1em] \Rightarrow x^3 - \frac{1}{x^3} = -24\sqrt{3} - 6\sqrt{3} \\[1em] \Rightarrow x^3 - \frac{1}{x^3} = -30\sqrt{3}.\end{aligned}$$

Hence, $x^3 - \frac{1}{x^3} = -30\sqrt{3}$.

Let the two numbers be x and y.

Given,

Sum of two numbers is 7.

∴ x + y = 7

Sum of cubes of two numbers is 133.

∴ x³ + y³ = 133

By formula,

⇒ (x + y)³ = x³ + y³ + 3xy(x + y)

Substituting values we get :

⇒ 7³ = 133 + 3xy × 7

⇒ 343 = 133 + 21xy

⇒ 21xy = 343 - 133

⇒ 21xy = 210

⇒ xy = $\frac{210}{21}$ = 10.

By formula,

⇒ (x + y)² = x² + y² + 2xy

Substituting values we get :

⇒ 7² = x² + y² + 2 × 10

⇒ 49 = x² + y² + 20

⇒ x² + y² = 49 - 20 = 29.

Hence, sum of the squares of the numbers = 29.

(i) Given,

a - b = 7, cubing both sides we get,

⇒ (a - b)³ = 7³

⇒ a³ - b³ - 3ab(a - b) = 343

⇒ 133 - 3ab(7) = 343

⇒ 133 - 21ab = 343

⇒ -21ab = 343 - 133

⇒ -21ab = 210

⇒ ab = -10.

Hence, ab = -10.

(ii) We know that,

a² + b² = (a - b)² + 2ab

Substituting values we get,

⇒ a² + b² = (7)² + 2(-10)

⇒ a² + b² = 49 - 20 = 29.

Hence, a² + b² = 29.

The above equation can be written as,

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

Hence, coefficient of x² = 6.

Given,

x² + y² + z² = xy + yz + zx

⇒ x² + y² + z² - (xy + yz + zx) = 0

⇒ x² + y² + z² - xy - yz - zx = 0

Multiplying by 2 on both sides,

⇒ 2x² + 2y² + 2z² - 2xy - 2yz - 2zx = 0

⇒ (x² + y² - 2xy) + (y² + z² - 2yz) + (z² + x² - 2xz) = 0

⇒ (x - y)² + (y - z)² + (z - x)² = 0

⇒ x - y = 0, y - z = 0 and z - x = 0

⇒ x = y, y = z and z = x

⇒ x = y = z

Hence, proved that if x² + y² + z² = xy + yz + zx, then x = y = z