ICSE Class 8 Mathematics · Selina Solutions Concise Mathematics Class 8 ICSE · 119 solved questions
(x + 2y)2 + (x - 2y)2 is equal to:
(x + 2y)2 + (x - 2y)2
And using the formula,
[∵(x + y)2 = x2 + 2xy + y2]
and,
[∵(x - y)2 = x2 - 2xy + y2]
= [x2 + 2 $\times$ x $\times$ 2y + (2y)2] + [x2 - 2 $\times$ x $\times$ 2y + (2y)2]
= [x2 + 4xy + 4y2] + [x2 - 4xy + 4y2]
= x2 + 4xy + 4y2 + x2 - 4xy + 4y2
= (x2 + x2) + (4xy - 4xy) + (4y2 + 4y2)
= 2x2 + 8y2
Hence, option 3 is the correct option.
(a + b) (a - b) + (b - c) (b + c) + (c + a) (c - a) is equal to:
[(a + b) (a - b)] + [(b - c) (b + c)] + [(c + a) (c - a)]
Using the formula,
[∵ (x + y)(x - y) = x2 - y2]
= [a2 - b2] + [b2 - c2] + [c2 - a2]
= a2 - b2 + b2 - c2 + c2 - a2
= (a2 - a2) + (- b2 + b2) + (- c2 + c2)
= 0
Hence, option 3 is the correct option.
(3x - 4y)2 - (3x + 4y)2 is equal to:
(3x - 4y)2 - (3x + 4y)2
Using the formula,
[∵(x + y)2 = x2 + 2xy + y2]
and,
[∵(x - y)2 = x2 - 2xy + y2]
= [(3x)2 - 2 $\times$ (3x) $\times$ (4y) + (4y)2] - [(3x)2 + 2 $\times$ (3x) $\times$ (4y) + (4y)2]
= [9x2 - 24xy + 16y2] - [9x2 + 24xy + 16y2]
= 9x2 - 24xy + 16y2 - 9x2 - 24xy - 16y2
= (9x2 - 9x2) + (- 24xy - 24xy) + (16y2 - 16y2)
= - 48xy
Hence, option 3 is the correct option.
The value of (0.8)2 - 0.32 + (0.2)2 is equal to:
(0.8)2 - 0.32 + (0.2)2
= (0.8)2 - 2 $\times$ (0.8) $\times$ (0.2) + (0.2)2
Using the formula,
[∵(x - y)2 = x2 - 2xy + y2]
= [0.8 - 0.2]2
= 0.62
= 0.36
Hence, option 3 is the correct option.
The value of (a - b - c)(a - b + c) is:
(a - b - c)(a - b + c)
= a (a - b + c) - b (a - b + c) - c (a - b + c)
= a(1+1) - ab + ac - ab + b(1+1) - bc -ac + bc - c(1+1)
= a2 + (- ab - ab) + (ac -ac) + b2 + (- bc + bc) - c2
= a2 - 2ab + b2 - c2
Hence, option 2 is the correct option.
Use direct method to evaluate the following product:
(a - 8) (a + 2)
(a - 8) (a + 2)
Using the formula,
(x - a)(x + b) = x2 - (a - b)x - ab
= a2 - (8 - 2)a - 8 $\times$ 2
= a2 - 6a - 16
Hence, (a - 8) (a + 2) = a2 - 6a - 16
Use direct method to evaluate the following product:
(b - 3) (b - 5)
(b - 3) (b - 5)
Using the formula,
(x + a)(x + b) = x2 - (a + b)x + ab
= b2 - (3 + 5)b + 3 $\times$ 5
= b2 - 8b + 15
Hence, (b - 3) (b - 5) = b2 - 8b + 15
Use direct method to evaluate the following product:
(3x - 2y) (2x + y)
(3x - 2y) (2x + y)
= (3x $\times$ 2x) + (3x $\times$ y + (-2y) $\times$ 2x) + ((-2y) $\times$ y)
= 6x2 + (3xy - 4xy) - 2y2
= 6x2 - 1xy - 2y2
Hence, (3x - 2y) (2x + y) = 6x2 - xy - 2y2
Use direct method to evaluate the following product:
(5a + 16) (3a - 7)
(5a + 16) (3a - 7)
= (5a $\times$ 3a) + (5a $\times$ (- 7) + 3a $\times$ 16) + (16 $\times$ (- 7))
= 15a2 + (-35a + 48a) - 112
= 15a2 + 13a - 112
Hence, (5a + 16) (3a - 7) = 15a2 + 13a - 112
Use direct method to evaluate the following product:
(8 - b) (3 + b)
(8 - b) (3 + b)
= (8 $\times$ 3) + (8 $\times$ b + (- b) $\times$ 3) + ((- b) $\times$ b)
= 24 + (8b - 3b) - b2
= 24 + 5b - b2
Hence, (8 - b) (3 + b) = 24 + 5b - b2
Evaluate:
(2a + 3) (2a - 3)
(2a + 3) (2a - 3)
Using the formula
[∵ (x + y)(x - y) = x2 - y2]
= (2a)2 - 32
= 4a2 - 9
Hence, (2a + 3) (2a - 3) = 4a2 - 9
Evaluate:
(xy + 4) (xy - 4)
(xy + 4) (xy - 4)
Using the formula
[∵ (x + y)(x - y) = x2 - y2]
= (xy)2 - 42
= x2y2 - 16
Hence, (xy + 4) (xy - 4) = x2y2 - 16
Evaluate:
(ab + x2) (ab - x2)
(ab + x2) (ab - x2)
Using the formula
[∵ (x + y)(x - y) = x2 - y2]
= (ab)2 - (x2)2
= a2b2 - x4
Hence, (ab + x2) (ab - x2) = a2b2 - x4
Evaluate:
(3x2 + 5y2) (3x2 - 5y2)
(3x2 + 5y2) (3x2 - 5y2)
Using the formula
[∵ (x + y)(x - y) = x2 - y2]
= (3x2)2 - (5y2)2
= 9x4 - 25y4
Hence, (3x2 + 5y2) (3x2 - 5y2) = 9x4 - 25y4
Evaluate:
$\Big(z -\frac{2}{3}\Big)$ $\Big(z +\frac{2}{3}\Big)$
$\Big(z -\frac{2}{3}\Big)$ $\Big(z +\frac{2}{3}\Big)$
Using the formula
[∵ (x + y)(x - y) = x2 - y2]
= z2 - $\Big(\frac{2}{3}\Big)$2
= z2 - $\frac{4}{9}$
Hence, $\Big(z -\frac{2}{3}\Big)$ $\Big(z +\frac{2}{3}\Big)$ = z2 - $\frac{4}{9}$
Evaluate:
$\Big(\frac{3}{5}a +\frac{1}{2}\Big)$ $\Big(\frac{3}{5}a -\frac{1}{2}\Big)$
$\Big(\frac{3}{5}a +\frac{1}{2}\Big)$ $\Big(\frac{3}{5}a -\frac{1}{2}\Big)$
Using the formula
[∵ (x + y)(x - y) = x2 - y2]
= $\Big(\frac{3}{5}a\Big)^2 - \Big(\frac{1}{2}\Big)^2$
= $\Big(\frac{9}{25}a^2\Big) - \frac{1}{4}$
Hence, $\Big(\frac{3}{5}a +\frac{1}{2}\Big)$ $\Big(\frac{3}{5}a -\frac{1}{2}\Big)$ = $\Big(\frac{9}{25}a^2\Big) - \frac{1}{4}$
Evaluate:
(0.5 - 2a) (0.5 + 2a)
(0.5 - 2a) (0.5 + 2a)
Using the formula
[∵ (x + y)(x - y) = x2 - y2]
= (0.5)2 - (2a)2
= 0.25 - 4a2
Hence, (0.5 - 2a) (0.5 + 2a) = 0.25 - 4a2
Evaluate:
$\Big(\frac{a}{2} -\frac{b}{3}\Big)$ $\Big(\frac{a}{2} +\frac{b}{3}\Big)$
$\Big(\frac{a}{2} -\frac{b}{3}\Big)$ $\Big(\frac{a}{2} +\frac{b}{3}\Big)$
Using the formula
[∵ (x + y)(x - y) = x2 - y2]
= $\Big(\frac{a}{2}\Big)^2 - \Big(\frac{b}{3}\Big)^2$
= $\Big(\frac{a^2}{4}\Big) - \Big(\frac{b^2}{9}\Big)$
Hence, $\Big(\frac{a}{2} -\frac{b}{3}\Big)$ $\Big(\frac{a}{2} +\frac{b}{3}\Big)$ = $\Big(\frac{a^2}{4}\Big) - \Big(\frac{b^2}{9}\Big)$
Evaluate:
(a + b) (a - b) (a2 + b2)
(a + b) (a - b) (a2 + b2)
Using the formula
[∵ (x + y)(x - y) = x2 - y2]
= (a2 - b2) (a2 + b2)
= a4 - b4
Hence, (a + b) (a - b) (a2 + b2) = a4 - b4
Evaluate:
(3 - 2x) (3 + 2x) (9 + 4x2)
(3 - 2x) (3 + 2x) (9 + 4x2)
Using the formula
[∵ (x + y)(x - y) = x2 - y2]
= (32 - (2x)2) (9 + 4x2)
= (9 - 4x2) (9 + 4x2)
= 92 - (4x2)2
= 81 - 16x4
Hence, (3 - 2x) (3 + 2x) (9 + 4x2) = 81 - 16x4
Evaluate:
(3x - 4y) (3x + 4y) (9x2 + 16y2)
(3x - 4y) (3x + 4y) (9x2 + 16y2)
Using the formula
[∵ (x + y)(x - y) = x2 - y2]
= [(3x)2 - (4y)2] (9x2 + 16y2)
= (9x2 - 16y2) (9x2 + 16y2)
= (9x2)2 - (16y2)2
= 81x4 - 256y4
Hence, (3x - 4y) (3x + 4y) (9x2 + 16y2) = 81x4 - 256y4
Use the formula: (a + b) (a - b) = a2 - b2 to evaluate:
(i) 21 x 19
(ii) 33 x 27
(iii) 103 x 97
(iv) 9.8 x 10.2
(v) 7.7 x 8.3
(i) 21 x 19
(20 + 1)(20 - 1)
Using the formula: (a + b) (a - b) = a2 - b2
⇒ (20 + 1)(20 - 1) = 202 - 12
= 400 - 1
= 399
Hence, 21 x 19 = 399.
(ii) 33 x 27
(30 + 3)(30 - 3)
Using the formula: (a + b) (a - b) = a2 - b2
⇒ (30 + 3)(30 - 3) = 302 - 32
= 900 - 9
= 891
Hence, 33 x 27 = 891.
(iii) 103 x 97
(100 + 3)(100 - 3)
Using the formula: (a + b) (a - b) = a2 - b2
⇒ (100 + 3)(100 - 3) = 1002 - 32
= 10000 - 9
= 9991
Hence, 103 x 97 = 9991.
(iv) 9.8 x 10.2
(10 - 0.2)(10 + 0.2)
Using the formula: (a + b) (a - b) = a2 - b2
⇒ (10 - 0.2)(10 + 0.2) = 102 - (0.2)2
= 100 - 0.04
= 99.96
Hence, 9.8 x 10.2 = 99.96.
(v) 7.7 x 8.3
(8 - 0.3)(8 + 0.3)
Using the formula: (a + b) (a - b) = a2 - b2
⇒ (8 - 0.3)(8 + 0.3) = 82 - (0.3)2
= 64 - 0.09
= 63.91
Hence, 7.7 x 8.3 = 63.91.
Evaluate:
(6 - xy) (6 + xy)
(6 - xy) (6 + xy)
Using the formula
[∵ (x + y)(x - y) = x2 - y2]
= 62 - (xy)2
= 36 - x2y2
Hence, (6 - xy) (6 + xy) = 36 - x2y2
Evaluate:
$\Big(7x +\frac{2}{3}y\Big)$ $\Big(7x -\frac{2}{3}y\Big)$
$\Big(7x +\frac{2}{3}y\Big)$ $\Big(7x -\frac{2}{3}y\Big)$
Using the formula
[∵ (x + y)(x - y) = x2 - y2]
$$\begin{aligned}= \Big(7x\Big)^2 +\Big(\frac{2}{3}y\Big)^2\\[1em] = 49x^2 + \frac{4}{9}y^2\end{aligned}$$
Hence, $\Big(7x +\frac{2}{3}y\Big)$ $\Big(7x -\frac{2}{3}y\Big)$ = $49x^2 - \Big(\frac{4}{9}y^2\Big)$
Evaluate:
$\Big(\frac{a}{2b} +\frac{2b}{a}\Big)$ $\Big(\frac{a}{2b} -\frac{2b}{a}\Big)$
$\Big(\frac{a}{2b} +\frac{2b}{a}\Big)$ $\Big(\frac{a}{2b} -\frac{2b}{a}\Big)$
Using the formula
[∵ (x + y)(x - y) = x2 - y2]
= $\Big(\frac{a}{2b}\Big)^2 - \Big(\frac{2b}{a}\Big)^2$
= $\Big(\frac{a^2}{4b^2}\Big) - \Big(\frac{4b^2}{a^2}\Big)$
Hence, $\Big(\frac{a}{2b} +\frac{2b}{a}\Big)$ $\Big(\frac{a}{2b} -\frac{2b}{a}\Big)$ = $\Big(\frac{a^2}{4b^2}\Big) - \Big(\frac{4b^2}{a^2}\Big)$
Evaluate:
$\Big(3x -\frac{1}{2y}\Big)$ $\Big(3x +\frac{1}{2y}\Big)$
$\Big(3x -\frac{1}{2y}\Big)$ $\Big(3x +\frac{1}{2y}\Big)$
Using the formula
[∵ (x + y)(x - y) = x2 - y2]
$$\begin{aligned}= (3x)^2 - \Big(\frac{1}{2y}\Big)^2\\[1em] = 9x^2 - \Big(\frac{1}{4y^2}\Big)\end{aligned}$$
Hence, $\Big(3x -\frac{1}{2y}\Big)$ $\Big(3x +\frac{1}{2y}\Big)$ = $9x^2 - \Big(\frac{1}{4y^2}\Big)$
Evaluate:
(2a + 3) (2a - 3) (4a2 + 9)
(2a + 3) (2a - 3) (4a2 + 9)
Using the formula
[∵ (x + y)(x - y) = x2 - y2]
= [(2a)2 - 32] (4a2 + 9)
= (4a2 - 92) (4a2 + 9)
= (4a2)2 - 92
= 16a4 - 81
Hence, (2a + 3) (2a - 3) (4a2 + 9) = 16a4 - 81
Evaluate:
(a + bc) (a - bc) (a2 + b2c2)
(a + bc) (a - bc) (a2 + b2c2)
Using the formula
[∵ (x + y)(x - y) = x2 - y2]
= [a2 - (bc)2] (a2 + b2c2)
= (a2 - b2c2) (a2 + b2c2)
= (a2)2 - (b2c2)2
= a4 - b4c4
Hence, (a + bc) (a - bc) (a2 + b2c2) = a4 - b4c4
Expand:
$\Big(a +\frac{1}{2a}\Big)^2$
$\Big(a +\frac{1}{2a}\Big)^2$
Using the formula:
(∵ (x + y)2 = x2 + 2xy + y2)
$$\begin{aligned}= (a)^2 + 2 \times a \times \frac{1}{2a} + \Big(\frac{1}{2a}\Big)^2\\[1em] = a^2 + \frac{2a}{2a} + \frac{1}{4a^2}\\[1em] = a^2 + 1 + \frac{1}{4a^2}\end{aligned}$$
Hence, $\Big(a +\frac{1}{2a}\Big)^2$ = $a^2 + 1 + \Big(\frac{1}{4a^2}\Big)$
Expand:
$\Big(2a -\frac{1}{a}\Big)^2$
$\Big(2a -\frac{1}{a}\Big)^2$
Using the formula:
(∵ (x + y)2 = x2 + 2xy + y2)
$$\begin{aligned}= (2a)^2 - 2 \times 2a \times \frac{1}{a} + \Big(\frac{1}{a}\Big)^2\\[1em] = 4a^2 - \frac{4a}{a} + \frac{1}{a^2}\\[1em] = 4a^2 - 4 + \frac{1}{a^2}\end{aligned}$$
Hence, $\Big(2a -\frac{1}{a}\Big)^2$ = $4a^2 - 4 + \Big(\frac{1}{a^2}\Big)$
Expand:
(a + b - c)2
(a + b - c)2
Using the formula,
[∵ (x + y - z)2 = x2 + y2 + z2 + 2xy - 2yz - 2xz]
= a2 + b2 + c2 + 2ab - 2bc - 2ca
Hence, (a + b - c)2 = a2 + b2 + c2 + 2ab - 2bc - 2ca
Expand:
(a - b + c)2
(a - b + c)2
Using the formula,
[∵ (x - y + z)2 = x2 + y2 + z2 - 2xy - 2yz + 2xz]
= a2 + b2 + c2 - 2ab - 2bc + 2ca
Hence, (a + b - c)2 = a2 + b2 + c2 - 2ab - 2bc + 2ca
Expand:
$\Big(3x +\frac{1}{3x}\Big)^2$
$\Big(3x +\frac{1}{3x}\Big)^2$
Using the formula
(∵ (x + y)2 = x2 + 2xy + y2)
$$\begin{aligned}= (3x)^2 + 2 \times 3x \times \frac{1}{3x} + \frac{1}{3x}^2\\[1em] = 9x^2 + \frac{6x}{3x} + \frac{1}{9x^2}\\[1em] = 9x^2 + 2 + \frac{1}{9x^2}\end{aligned}$$ Hence, $\Big(3x +\frac{1}{3x}\Big)^2 = 9x^2 + 2 + \frac{1}{9x^2}$
Find the square of:
$a +\frac{1}{5a}$
$(a +\frac{1}{5a})^2$
Using the formula
(∵ (x + y)2 = x2 + 2xy + y2)
$$\begin{aligned}= a^2 + 2 \times a \times \frac{1}{5a} + \frac{1}{5a}^2\\[1em] = a^2 + \frac{2a}{5a} + \frac{1}{25a^2}\\[1em] = a^2 + \frac{2}{5} + \frac{1}{25a^2}\end{aligned}$$
Hence, $(a +\frac{1}{5a})^2$ = a2 + $\frac{2}{5}$ + $\frac{1}{25a^2}$
Find the square of:
$2a -\frac{1}{a}$
$(2a - \frac{1}{a})^2$
Using the formula
(∵ (x - y)2 = x2 - 2xy + y2)
$$\begin{aligned}= (2a)^2 - 2 \times 2a \times \frac{1}{a} + \frac{1}{a}^2\\[1em] = 4a^2 - \frac{4a}{a} + \frac{1}{a^2}\\[1em] = 4a^2 - 4 + \frac{1}{a^2}\end{aligned}$$
Hence, $(2a - \frac{1}{a})^2$ = $4a^2 - 4 + \frac{1}{a^2}$
Find the square of:
x - 2y + 1
(x - 2y + 1)2
Using the formula
(∵ (x - y + z)2 = x2 + y2 + z2 - 2xy - 2yz + 2xz)
= x2 + (2y)2 + 12 - 2 $\times$ (x) $\times$ (2y) - 2 $\times$ (2y) $\times$ 1 + 2 $\times$ 1 $\times$ (x)
= x2 + 4y2 + 1 - 4xy - 4y + 2x
Hence,(x - 2y + 1)2 = x2 + 4y2 + 1 - 4xy - 4y + 2x
Find the square of:
3a - 2b - 5c
(3a - 2b - 5c)2
Using the formula
(∵ (x - y - z)2 = x2 + y2 + z2 - 2xy + 2yz - 2xz)
= (3a)2 + (2b)2 + (5c)2 - 2 $\times$ (3a) $\times$ (2b) - 2 $\times$ (2b) $\times$ (5c) + 2 $\times$ (5c) $\times$ (3a)
= 9a2 + 4b2 + 25c2 - 12ab + 20bc - 30ca
Hence, (3a - 2b - 5c)2 = 9a2 + 4b2 + 25c2 - 12ab + 20bc - 30ca
Find the square of:
$2x +\frac{1}{x}+ 1$
$(2x +\frac{1}{x}+ 1)$2
Using the formula
(∵ (x + y + z)2 = x2 + y2 + z2 + 2xy + 2yz + 2xz)
$$\begin{aligned}= (2x)^2 + \Big(\frac{1}{x}\Big)^2 + 1^2 + 2 \times 2x \times \Big(\frac{1}{x}\Big) + 2 \times \Big(\frac{1}{x}\Big) \times 1 + 2 \times 1 \times 2x\\[1em] = 4x^2 + \Big(\frac{1}{x^2}\Big) + 1 + \Big(\frac{4x}{x}\Big) + \Big(\frac{2}{x}\Big) + 4x\\[1em] = 4x^2 + \Big(\frac{1}{x^2}\Big) + 1 + 4 + \Big(\frac{2}{x}\Big) + 4x\\[1em] = 4x^2 + \Big(\frac{1}{x^2}\Big) + 5 + \Big(\frac{2}{x}\Big) + 4x\end{aligned}$$
Hence, $(2x +\frac{1}{x}+ 1)$2 = 4x2 + $\Big(\frac{1}{x^2}\Big)$ + 5 + $\Big(\frac{2}{x}\Big)$ + 4x
Find the square of:
$5 - x +\frac{2}{x}$
$(5 - x +\frac{2}{x})^2$
Using the formula
(∵ (x - y + z)2 = x2 + y2 + z2 - 2xy - 2yz + 2xz)
$$\begin{aligned}= (5)^2 + (x)^2 + \Big(\frac{2}{x}\Big)^2 - 2 \times 5 \times x - 2 \times x \times \Big(\frac{2}{x}\Big) + 2 \times \Big(\frac{2}{x}\Big) \times 5\\[1em] = 25 + x^2 + \Big(\frac{4}{x^2}\Big) - 10x - \Big(\frac{4x}{x}\Big) + \Big(\frac{20}{x}\Big)\\[1em] = 25 + x^2 + \Big(\frac{4}{x^2}\Big) - 10x - 4 + \Big(\frac{20}{x}\Big)\\[1em] = (25 - 4) + x^2 + \Big(\frac{4}{x^2}\Big) - 10x + \Big(\frac{20}{x}\Big)\\[1em] = 21 + x^2 + \Big(\frac{4}{x^2}\Big) - 10x + \Big(\frac{20}{x}\Big)\end{aligned}$$
Hence, $(5 - x +\frac{2}{x})^2$ = 21 + x2 + $\Big(\frac{4}{x^2}\Big)$ - 10x + $\Big(\frac{20}{x}\Big)$
Find the square of:
2x - 3y + z
(2x - 3y + z)2
Using the formula
(∵ (x - y + z)2 = x2 + y2 + z2 - 2xy - 2yz + 2xz)
= (2x)2 + (3y)2 + (z)2 - 2 $\times$ (2x) $\times$ (3y) - 2 $\times$ (3y) $\times$ (z) + 2 $\times$ (z) $\times$ (2x)
= 4x2 + 9y2 + z2 - 12xy - 6yz + 4xz
Hence, (2x - 3y + z)2 = 4x2 + 9y2 + z2 - 12xy - 6yz + 4xz
Find the square of:
$x +\frac{1}{x}- 1$
$(x +\frac{1}{x}- 1)^2$
Using the formula
(∵ (x + y - z)2 = x2 + y2 + z2 + 2xy - 2yz - 2xz)
$$\begin{aligned}= (x)^2 + \Big(\frac{1}{x}\Big)^2 + (1)^2 + 2 \times x \times \Big(\frac{1}{x}\Big) - 2 \times \Big(\frac{1}{x}\Big) \times 1 - 2 \times 1 \times x\\[1em] = x^2 + \Big(\frac{1}{x^2}\Big) + 1 + \Big(\frac{2x}{x}\Big) - \Big(\frac{2}{x}\Big) - 2x\\[1em] = x^2 + \Big(\frac{1}{x^2}\Big) + 1 + 2 - \Big(\frac{2}{x}\Big) - 2x\\[1em] = x^2 + \Big(\frac{1}{x^2}\Big) + 3 - \Big(\frac{2}{x}\Big) - 2x\end{aligned}$$
Hence, $(x +\frac{1}{x}- 1)^2$ = x2 + $\Big(\frac{1}{x^2}\Big)$ + 3 - $\Big(\frac{2}{x}\Big)$ - 2x
Evaluate using expansion of (a + b)2 or (a - b)2 :
(i) (208)2
(ii) (92)2
(iii) (9.4)2
(iv) (20.7)2
(i) (208)2
(200 + 8)2
Using the formula
(∵ (x + y)2 = x2 + 2xy + y2)
= (200)2 + 2 x 200 x 8 + (8)2
= 40000 + 3200 + 64
= 43264
Hence, (208)2 = 43264
(ii) (92)2
(100 - 8)2
Using the formula
(∵ (x - y)2 = x2 - 2xy + y2)
= (100)2 - 2 x 100 x 8 + (8)2
= 10000 - 1600 + 64
= 8400 + 64
= 8464
Hence, (92)2 = 8464
(iii) (9.4)2
(10 - 0.6)2
Using the formula
(∵ (x - y)2 = x2 - 2xy + y2)
= (10)2 - 2 x 10 x 0.6 + (0.6)2
= 100 - 12 + 0.36
= 88 + 0.36
= 88.36
Hence, (9.4)2 = 88.36
(iv) (20.7)2
(20 + 0.7)2
Using the formula
(∵ (x + y)2 = x2 + 2xy + y2)
= (20)2 + 2 x 20 x 0.7 + (0.7)2
= 400 + 28 + 0.49
= 428 + 0.49
= 428.49
Hence, (20.7)2 = 428.49
Expand:
(2a + b)3
(2a + b)3
Using the formula,
(x + y)3 = x3 + y3 + 3x2y + 3xy2
= (2a)3 + b3 + 3 x (2a)2x (b) + 3 x (2a) x (b)2
= 8a3 + b3 + 12a2b + 6ab2
Hence, (2a + b)3 = 8a3 + b3 + 12a2b + 6ab2
Expand:
(a - 2b)3
(a - 2b)3
Using the formula,
(x - y)3 = x3 - y3 - 3x2y + 3xy2
= a3 - (2b)3 - 3 x (a)2x (2b) + 3 x (a) x (2b)2
= a3 - 8b3 - 6a2b + 12ab2
Hence, (a - 2b)3 = a3 - 8b3 - 6a2b + 12ab2
Expand:
(3x - 2y)3
(3x - 2y)3
Using the formula,
(x - y)3 = x3 - y3 - 3x2y + 3xy2
= (3x)3 - (2y)3 - 3 x (3x)2x (2y) + 3 x (3x) x (2y)2
= 27x3 - 8y3 - 54x2y + 36xy2
Hence, (3x - 2y)3 = 27x3 - 8y3 - 54x2y + 36xy2
Expand:
(x + 5y)3
(x + 5y)3
Using the formula,
(x + y)3 = x3 + y3 + 3x2y + 3xy2
= (x)3 + (5y)3 + 3 $\times$ (x)2 $\times$ (5y) + 3 $\times$ (x) $\times$ (5y)2
= x3 + 125y3 + 15x2y + 75xy2
Hence, (x + 5y)3 = x3 + 125y3 + 15x2y + 75xy2
Expand:
$\Big(a +\frac{1}{a}\Big)^3$
$\Big(a +\frac{1}{a}\Big)^3$
Using the formula,
(x + y)3 = x3 + y3 + 3x2y + 3xy2
$$\begin{aligned}= (a)^3 + \Big(\frac{1}{a}\Big)^3 + 3 \times (a)^2 \times \Big(\frac{1}{a}\Big) + 3 \times (a) \times \Big(\frac{1}{a}\Big)^2\\[1em] = a^3 + \Big(\frac{1}{a^3}\Big) + \Big(\frac{3a^2}{a}\Big) + \Big(\frac{3a}{a^2}\Big)\\[1em] = a^3 + \Big(\frac{1}{a^3}\Big) + 3a + \Big(\frac{3}{a}\Big)\\[1em]\end{aligned}$$
Hence, $\Big(a +\frac{1}{a}\Big)^3 = a^3 + \Big(\frac{1}{a^3}\Big) + 3a + \Big(\frac{3}{a}\Big)$
Expand:
$\Big(2a -\frac{1}{2a}\Big)^3$
$\Big(2a -\frac{1}{2a}\Big)^3$
Using the formula,
(x - y)3 = x3 - y3 - 3x2y + 3xy2
$$\begin{aligned}= (2a)^3 - \Big(\frac{1}{2a}\Big)^3 - 3 \times (2a)^2 \times \Big(\frac{1}{2a}\Big) + 3 \times (2a) \times \Big(\frac{1}{2a}\Big)^2\\[1em] = 8a^3 - \Big(\frac{1}{8a^3}\Big) - \Big(\frac{12a^2}{2a}\Big) + \Big(\frac{6a}{4a^2}\Big)\\[1em] = 8a^3 - \Big(\frac{1}{8a^3}\Big) - 6a + \Big(\frac{3}{2a}\Big)\\[1em]\end{aligned}$$
Hence, $\Big(2a - \frac{1}{2a}\Big)^3 = 8a^3 - \Big(\frac{1}{8a^3}\Big) - 6a + \Big(\frac{3}{2a}\Big)$
Find the cube of:
a + 2
(a + 2)3
Using the formula,
(x + y)3 = x3 + y3 + 3x2y + 3xy2
= (a)3 + (2)3 + 3 $\times$ (a)2 $\times$ (2) + 3 $\times$ (a) $\times$ (2)2
= a3 + 8 + 6a2 + 12a
Hence, (a + 2)3 = a3 + 8 + 6a2 + 12a
Find the cube of:
2a - 1
(2a - 1)3
Using the formula,
(x - y)3 = x3 - y3 - 3x2y + 3xy2
= (2a)3 - (1)3 - 3 $\times$ (2a)2 $\times$ (1) + 3 $\times$ (2a) $\times$ (1)2
= 8a3 - 1 - 12a2 + 6a
Hence, (2a - 1)3 = 8a3 - 1 - 12a2 + 6a
Find the cube of:
2a + 3b
(2a + 3b)3
Using the formula,
(x + y)3 = x3 + y3 + 3x2y + 3xy2
= (2a)3 + (3b)3 + 3 $\times$ (2a)2 $\times$ (3b) + 3 $\times$ (2a) $\times$ (3b)2
= 8a3 + 27b3 + 36a2b + 54ab2
Hence, (2a + 3b)3 = 8a3 + 27b3 + 36a2b + 54ab2
Find the cube of:
3b - 2a
(3b - 2a)3
Using the formula,
(x - y)3 = x3 - y3 - 3x2y + 3xy2
= (3b)3 - (2a)3 + 3 $\times$ (3b)2 $\times$ (2a) + 3 $\times$ (3b) $\times$ (2a)2
= 27b3 - 8a3 - 54b2a + 36ba2
Hence, (3b - 2a)3 = 27b3 - 8a3 - 54b2a + 36ba2
Find the cube of:
$2x +\frac{1}{x}$
$2x + \frac{1}{x}$
Using the formula,
(x + y)3 = x3 + y3 + 3x2y + 3xy2
$$\begin{aligned}= (2x)^3 + \Big(\frac{1}{x}\Big)^3 + 3 \times (2x)^2 \times \Big(\frac{1}{x}\Big) + 3 \times 2x \times \Big(\frac{1}{x}\Big)^2\\[1em] = 8x^3 + \Big(\frac{1}{x^3}\Big) + \Big(\frac{12x^2}{x}\Big) + \Big(\frac{6x}{x^2}\Big)\\[1em] = 8x^3 + \Big(\frac{1}{x^3}\Big) + 12x + \frac{6}{x}\\[1em]\end{aligned}$$
Hence, $2x + \frac{1}{x} = 8x^3 + \Big(\frac{1}{x^3}\Big) + 12x + \frac{6}{x}$
Find the cube of:
$x -\frac{1}{2}$
$x - \frac{1}{2}$
Using the formula,
(x - y)3 = x3 - y3 - 3x2y + 3xy2
$$\begin{aligned}= (x)^3 - \Big(\frac{1}{2}\Big)^3 - 3 \times x^2 \times \Big(\frac{1}{2}\Big) + 3 \times x \times \Big(\frac{1}{2}\Big)^2\\[1em] = x^3 - \Big(\frac{1}{8}\Big) - \Big(\frac{3x^2}{2}\Big) + \Big(\frac{3x}{4}\Big)\end{aligned}$$
Hence, $x + \frac{1}{2} = x^3 - \Big(\frac{1}{8}\Big) - \Big(\frac{3x^2}{2}\Big) + \Big(\frac{3x}{4}\Big)$
If $x -\frac{1}{x}= 3$, the value of $x^2 +\frac{1}{x^2}$ is:
Using the formula,
[∵(x - y)2 = x2 - 2xy + y2]
$$\begin{aligned}\Big(x -\frac{1}{x}\Big)^2 = x^2 - 2 \times x \times \frac{1}{x} + \Big(\frac{1}{x}\Big)^2\\[1em] = x^2 - \frac{2x}{x} + \Big(\frac{1}{x}\Big)^2\\[1em] = x^2 - 2 + \frac{1}{x^2}\\[1em]\end{aligned}$$
Putting the value, $x -\frac{1}{x}= 3$
$$\begin{aligned}(3)^2 = x^2 - 2 + \frac{1}{x^2}\\[1em] ⇒ 9 = x^2 - 2 + \frac{1}{x^2}\\[1em] ⇒ x^2 + \frac{1}{x^2} = 9 + 2\\[1em] ⇒ x^2 + \frac{1}{x^2} = 11\end{aligned}$$
Hence, option 3 is the correct option.
If a + b = 7 and ab = 10; the value of a2 + b2 is equal to:
Using the formula,
[∵(x + y)2 = x2 + 2xy + y2]
So,
(a + b)2 = a2 + 2ab + b2
Putting the value, a + b = 7 and ab = 10
⇒ (7)2 = a2 + 2 x 10 + b2
⇒ 49 = a2 + 20 + b2
⇒ a2 + b2 = 49 - 20
⇒ a2 + b2 = 29
Hence, option 1 is the correct option.
The value of $\frac{95 \times 95 - 5 \times 5}{95 - 5}$ is equal to:
Using the formula,
[∵(x2 - y2) = (x + y)(x + y)]
$$\begin{aligned}\frac{95 \times 95 - 5 \times 5}{95 - 5}\\[1em] \frac{(95)^2 - (5)^2}{95 - 5}\\[1em] = \frac{(95 - 5)(95 + 5)}{95 - 5}\\[1em] = \frac{\cancel{(95 - 5)}(95 + 5)}{\cancel{(95 - 5)}}\\[1em] = 95 + 5\\[1em] = 100\end{aligned}$$
Hence, option 1 is the correct option.
If a - b = 1 and a + b = 3, the value of ab is:
Using the formula,
[∵(x + y)2 = x2 + 2xy + y2]
And
[∵(x - y)2 = x2 - 2xy + y2]
(a + b)2 = a2 + 2ab + b2
Putting a + b = 3, we get
(3)2 = a2 + 2ab + b2
9 = a2 + 2ab + b2
9 - 2ab = a2 + b2 ...............(1)
And,
(a - b)2 = a2 - 2ab + b2
Putting the value, a - b = 1
(1)2 = a2 - 2ab + b2
1 = a2 + b2 - 2ab
Using equation (1)
1 = 9 - 2ab - 2ab
1 = 9 - 4ab
4ab = 9 - 1
4ab = 8
ab = $\frac{8}{4}$
ab = 2
Hence, option 2 is the correct option.
If $x +\frac{1}{x} = 2$, the value of $\Big(x^3 +\frac{1}{x^3}\Big) - \Big(x^2 +\frac{1}{x^2}\Big)$ is:
Using the formula,
[∵ (x + y)3 = x3 + 3xy(x + y) + y3]
And,
[∵ (x + y)2 = x2 + 2xy + y2]
$$\begin{aligned}\Big(x + \frac{1}{x}\Big)^3 = x^3 + 3 \times x \times \frac{1}{x}\Big(x + \frac{1}{x}\Big) + \Big(\frac{1}{x}\Big)^3\\[1em] ⇒ \Big(x + \frac{1}{x}\Big)^3 = x^3 + 3\Big(x + \frac{1}{x}\Big) + \frac{1}{x^3}\end{aligned}$$
Putting $x +\frac{1}{x} = 2$
$$\begin{aligned}2^3 = x^3 + 3 \times 2 + \frac{1}{x^3}\\[1em] ⇒ 8 = x^3 + 6 + \frac{1}{x^3}\\[1em] ⇒ x^3 + \frac{1}{x^3} = 8 - 6\\[1em] ⇒ x^3 + \frac{1}{x^3} = 2...............(1)\end{aligned}$$
And ,
$$\begin{aligned}\Big(x + \frac{1}{x}\Big)^2 = x^2 + 2 \times x \times \frac{1}{x} + \Big(\frac{1}{x}\Big)^2\\[1em] ⇒ \Big(x + \frac{1}{x}\Big)^2 = x^2 + 2 + \frac{1}{x^2}\end{aligned}$$
Putting the value $x +\frac{1}{x} = 2$,we get
$$\begin{aligned}2^2 = x^2 + 2 + \frac{1}{x^2}\\[1em] ⇒ 4 = x^2 + 2 + \frac{1}{x^2}\\[1em] ⇒ 4 - 2 = x^2 + \frac{1}{x^2}\\[1em] ⇒ 2 = x^2 + \frac{1}{x^2} ...............(2)\end{aligned}$$
Now, using equation (1) and (2),
$$\begin{aligned}\Big(x^3 +\frac{1}{x^3}\Big) - \Big(x^2 +\frac{1}{x^2}\Big)\\[1em] = 2 - 2\\[1em] = 0\end{aligned}$$
Hence, option 1 is the correct option.
If a + b = 5 and ab = 6, find a2 + b2
Using the formula,
[∵(x + y)2 = x2 + 2xy + y2]
So,
(a + b)2 = a2 + 2ab + b2
Putting the value, a + b = 5 and ab = 6
⇒ (5)2 = a2 + 2 x 6 + b2
⇒ 25 = a2 + 12 + b2
⇒ a2 + b2 = 25 - 12
⇒ a2 + b2 = 13
Hence, the value of (a2 + b2) is 13.
If a - b = 6 and ab = 16, find a2 + b2
Using the formula,
[∵(x - y)2 = x2 - 2xy + y2]
So,
(a - b)2 = a2 - 2ab + b2
Putting the value, a - b = 6 and ab = 16
⇒ (6)2 = a2 - 2 x 16 + b2
⇒ 36 = a2 - 32 + b2
⇒ a2 + b2 = 36 + 32
⇒ a2 + b2 = 68
Hence, the value of (a2 + b2) is 68.
If a2 + b2 = 29 and ab = 10, find:
(i) a + b
(ii) a - b
(i) Using the formula,
[∵(x + y)2 = x2 + 2xy + y2]
So,
(a + b)2 = a2 + 2ab + b2
Putting the value, a2 + b2 = 29 and ab = 10
⇒ (a + b)2 = (a2 + b2) + 2ab
⇒ (a + b)2 = (29) + 2 $\times$ 10
⇒ (a + b)2 = 29 + 20
⇒ (a + b)2 = 49
⇒ a + b = $\sqrt{49}$
⇒ a + b = 7 or -7
Hence, the values of (a + b) are 7 or -7.
(ii) Using the formula,
[∵(x - y)2 = x2 - 2xy + y2]
So,
(a - b)2 = a2 - 2ab + b2
Putting the value, a2 + b2 = 29 and ab = 10
⇒ (a - b)2 = (a2 + b2) - 2ab
⇒ (a - b)2 = (29) - 2 $\times$ 10
⇒ (a - b)2 = 29 - 20
⇒ (a - b)2 = 9
⇒ a - b = $\sqrt{9}$
⇒ a - b = 3 or -3
Hence, the values of (a - b) are 3 or -3.
If a2 + b2 = 10 and ab = 3, find:
(i) a - b
(ii) a + b
(i) Using the formula,
[∵(x - y)2 = x2 - 2xy + y2]
So,
(a - b)2 = a2 - 2ab + b2
Putting the value, a2 + b2 = 10 and ab = 3
⇒ (a - b)2 = (a2 + b2) - 2ab
⇒ (a - b)2 = (10) - 2 $\times$ 3
⇒ (a - b)2 = 10 - 6
⇒ (a - b)2 = 4
⇒ a - b = $\sqrt{4}$
⇒ a - b = 2 or -2
Hence, the values of (a - b) are 2 or -2.
(ii) Using the formula,
[∵(x + y)2 = x2 + 2xy + y2]
So,
(a + b)2 = a2 + 2ab + b2
Putting the value, a2 + b2 = 10 and ab = 3
⇒ (a + b)2 = (a2 + b2) + 2ab
⇒ (a + b)2 = (10) + 2 $\times$ 3
⇒ (a + b)2 = 10 + 6
⇒ (a + b)2 = 16
⇒ a + b = $\sqrt{16}$
⇒ a + b = 4 or -4
Hence, the values of (a + b) are 4 or -4.
If $a +\frac{1}{a}= 3$, find: $a^2 +\frac{1}{a^2}$
Using the formula,
[∵ (x + y)2 = x2 + 2xy + y2]
So,
$$\begin{aligned}\Big(a + \frac{1}{a}\Big)^2 = a^2 + 2 \times a \times \frac{1}{a} + \Big(\frac{1}{a}\Big)^2\\[1em] ⇒ \Big(a + \frac{1}{a}\Big)^2 = a^2 + 2 + \frac{1}{a^2}\end{aligned}$$
Putting the value $a + \frac{1}{a} = 3$,we get
$$\begin{aligned}3^2 = a^2 + 2 + \frac{1}{a^2}\\[1em] ⇒ 9 = a^2 + 2 + \frac{1}{a^2}\\[1em] ⇒ a^2 + \frac{1}{a^2} = 9 - 2 \\[1em] ⇒ a^2 + \frac{1}{a^2} = 7\end{aligned}$$
Hence, the value of $a^2 + \frac{1}{a^2}$ is 7.
If $a -\frac{1}{a}= 4$, find: $a^2 +\frac{1}{a^2}$
Using the formula,
[∵ (x - y)2 = x2 - 2xy + y2]
So,
$$\begin{aligned}\Big(a - \frac{1}{a}\Big)^2 = a^2 - 2 \times a \times \frac{1}{a} + \Big(\frac{1}{a}\Big)^2\\[1em] ⇒ \Big(a - \frac{1}{a}\Big)^2 = a^2 - 2 + \frac{1}{a^2}\end{aligned}$$
Putting the value $a - \frac{1}{a} = 4$,we get
$$\begin{aligned}4^2 = a^2 - 2 + \frac{1}{a^2}\\[1em] ⇒ 16 = a^2 - 2 + \frac{1}{a^2}\\[1em] ⇒ a^2 + \frac{1}{a^2} = 16 + 2\\[1em] ⇒ a^2 + \frac{1}{a^2} = 18\end{aligned}$$
Hence, the value of $a^2 + \frac{1}{a^2}$ is 18.
If $a^2 +\frac{1}{a^2}= 23$, find: $a +\frac{1}{a}$
Using the formula,
[∵ (x + y)2 = x2 + 2xy + y2]
So,
$$\begin{aligned}\Big(a + \frac{1}{a}\Big)^2 = a^2 + 2 \times a \times \frac{1}{a} + \Big(\frac{1}{a}\Big)^2\\[1em] ⇒ \Big(a + \frac{1}{a}\Big)^2 = a^2 + 2 + \frac{1}{a^2}\end{aligned}$$
Putting the value $a^2 +\frac{1}{a^2} = 23$,we get
$$\begin{aligned}⇒ \Big(a + \frac{1}{a}\Big)^2 = \Big(a^2 + \frac{1}{a^2}\Big) + 2\\[1em] ⇒ \Big(a + \frac{1}{a}\Big)^2 = 23 + 2\\[1em] ⇒ \Big(a + \frac{1}{a}\Big)^2 = 25\\[1em] ⇒ \Big(a + \frac{1}{a}\Big) = \sqrt{25}\\[1em] ⇒ \Big(a + \frac{1}{a}\Big) = 5 \text{ or} -5\end{aligned}$$
Hence, the values of $\Big(a + \frac{1}{a}\Big)$ are 5 or -5.
If $a^2 +\frac{1}{a^2}= 11$, find: $a -\frac{1}{a}$
Using the formula,
[∵ (x - y)2 = x2 - 2xy + y2]
So,
$$\begin{aligned}\Big(a - \frac{1}{a}\Big)^2 = a^2 - 2 \times a \times \frac{1}{a} + \Big(\frac{1}{a}\Big)^2\\[1em] ⇒ \Big(a - \frac{1}{a}\Big)^2 = a^2 - 2 + \frac{1}{a^2}\end{aligned}$$
Putting the value $a^2 + \frac{1}{a}^2 = 11$,we get
$$\begin{aligned}⇒ \Big(a - \frac{1}{a}\Big)^2 = \Big(a^2 + \frac{1}{a^2}\Big) - 2\\[1em] ⇒ \Big(a - \frac{1}{a}\Big)^2 = 11 - 2\\[1em] ⇒ \Big(a - \frac{1}{a}\Big)^2 = 9\\[1em] ⇒ \Big(a - \frac{1}{a}\Big) = \sqrt{9}\\[1em] ⇒ \Big(a - \frac{1}{a}\Big) = 3 \text{ or} -3\end{aligned}$$
Hence, the values of $\Big(a - \frac{1}{a}\Big)$ are 3 or -3.
a + b + c = 10 and a2 + b2 + c2 = 38, find ab + bc + ca
Using the formula,
[∵ (x + y + z)2 = x2 + y2 + z2 + 2xy + 2yz + 2zx]
So,
⇒ (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca
⇒ (a + b + c)2 = a2 + b2 + c2 + 2(ab + bc + ca)
Putting (a + b + c) = 10 and (a2 + b2 + c2) = 38, we get
⇒ (10)2 = 38 + 2(ab + bc + ca)
⇒ 100 = 38 + 2(ab + bc + ca)
⇒ 2(ab + bc + ca) = 100 - 38
⇒ 2(ab + bc + ca) = 62
⇒ ab + bc + ca = $\frac{62}{2}$
⇒ ab + bc + ca = 31
Hence, the value of (ab + bc + ca) is 31.
Find : a2 + b2 + c2, if a + b + c = 9 and ab + bc + ca = 24.
Using the formula,
[∵ (x + y + z)2 = x2 + y2 + z2 + 2xy + 2yz + 2zx]
So,
⇒ (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca
⇒ (a + b + c)2 = a2 + b2 + c2 + 2(ab + bc + ca)
Putting (a + b + c) = 9 and (ab + bc + ca) = 24, we get
⇒ (9)2 = a2 + b2 + c2 + 2 x 24
⇒ 81 = a2 + b2 + c2 + 48
⇒ a2 + b2 + c2 = 81 - 48
⇒ a2 + b2 + c2 = 33
Hence, the value of a2 + b2 + c2 is 33.
Find: a + b + c, if a2 + b2 + c2 = 83 and ab + bc + ca = 71
Using the formula,
[∵ (x + y + z)2 = x2 + y2 + z2 + 2xy + 2yz + 2zx]
So,
⇒ (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca
⇒ (a + b + c)2 = a2 + b2 + c2 + 2(ab + bc + ca)
Putting (a2 + b2 + c2) = 83 and (ab + bc + ca) = 71, we get
⇒ (a + b + c)2 = 83 + 2 x 71
⇒ (a + b + c)2 = 83 + 142
⇒ (a + b + c)2 = 225
⇒ a + b + c = $\sqrt{225}$
⇒ a + b + c = 15 or - 15
Hence, the values of (a + b + c) are 15 or - 15.
If a + b = 6 and ab = 8, find: a3 + b3.
Using the formula,
[∵ (x + y)3 = x3 + y3 + 3xy(x + y)]
So,
(a + b)3 = a3 + b3 + 3ab(a + b)
Putting the values a + b = 6 and ab = 8, we get
⇒ (6)3 = a3 + b3 + 3 x 8 x 6
⇒ 216 = a3 + b3 + 144
⇒ a3 + b3 = 216 - 144
⇒ a3 + b3 = 72
Hence, the value of a3 + b3 is 72.
If a - b = 3 and ab = 10, find: a3 - b3.
Using the formula,
[∵ (x - y)3 = x3 - y3 - 3xy(x - y)]
So,
(a - b)3 = a3 - b3 - 3ab(a - b)
Putting the values a - b = 3 and ab = 10, we get
⇒ (3)3 = a3 - b3 - 3 x 10 x 3
⇒ 27 = a3 + b3 - 90
⇒ a3 + b3 = 27 + 90
⇒ a3 + b3 = 117
Hence, the value of a3 + b3 is 117.
Find : $a^3 +\frac{1}{a^3}$, if $a +\frac{1}{a}= 5$.
Using the formula,
[∵ (x + y)3 = x3 + 3xy(x + y) + y3]
So,
$$\begin{aligned}\Big(a + \frac{1}{a}\Big)^3 = a^3 + 3 \times a \times \frac{1}{a}\Big(a + \frac{1}{a}\Big) + \Big(\frac{1}{a}\Big)^3\\[1em] ⇒ \Big(a + \frac{1}{a}\Big)^3 = a^3 + 3\Big(a + \frac{1}{a}\Big) + \frac{1}{a^3}\end{aligned}$$
Putting $a +\frac{1}{a} = 5$
$$\begin{aligned}5^3 = a^3 + 3 \times 5 + \frac{1}{a^3}\\[1em] ⇒ 125 = a^3 + 15 + \frac{1}{a^3}\\[1em] ⇒ a^3 + \frac{1}{a^3} = 125 - 15 \\[1em] ⇒ a^3 + \frac{1}{a^3} = 110\end{aligned}$$
Hence, the value of $a^3 + \frac{1}{a^3}$ is 110.
Find : $a^3 -\frac{1}{a^3}$, if $a -\frac{1}{a}= 4$.
Using the formula,
[∵ (x - y)3 = x3 - 3xy(x - y) - y3]
So,
$$\begin{aligned}\Big(a - \frac{1}{a}\Big)^3 = a^3 - 3 \times a \times \frac{1}{a}\Big(a - \frac{1}{a}\Big) - \Big(\frac{1}{a}\Big)^3\\[1em] ⇒ \Big(a - \frac{1}{a}\Big)^3 = a^3 - 3\Big(a - \frac{1}{a}\Big) - \frac{1}{a^3}\end{aligned}$$
Putting $a - \frac{1}{a} = 4$
$$\begin{aligned}4^3 = a^3 - 3 \times 4 - \frac{1}{a^3}\\[1em] ⇒ 64 = a^3 - 12 - \frac{1}{a^3}\\[1em] ⇒ a^3 - \frac{1}{a^3} = 64 + 12\\[1em] ⇒ a^3 - \frac{1}{a^3} = 76\end{aligned}$$
Hence, the value of $a^3 - \frac{1}{a^3}$ is 76.
If $2x -\frac{1}{2x} = 4$, find:
(i) $4x^2 +\frac{1}{4x^2}$
(ii) $8x^3 -\frac{1}{8x^3}$
(i) Using the formula,
[∵ (x - y)2 = x2 - 2xy + y2]
So,
$$\begin{aligned}\Big(2x - \frac{1}{2x}\Big)^2 = (2x)^2 - 2 \times 2x \times \frac{1}{2x} + \Big(\frac{1}{2x}\Big)^2\\[1em] ⇒ \Big(2x - \frac{1}{2x}\Big)^2 = 4x^2 - 2 + \frac{1}{4x^2}\end{aligned}$$
Putting the value $2x - \frac{1}{2x} = 4$,we get
$$\begin{aligned}4^2 = 4x^2 - 2 + \frac{1}{4x^2}\\[1em] ⇒ 16 = 4x^2 - 2 + \frac{1}{4x^2}\\[1em] ⇒ 4x^2 + \frac{1}{4x^2} = 16 + 2 \\[1em] ⇒ 4x^2 + \frac{1}{4x^2} = 18\end{aligned}$$
Hence, the value of $4x^2 + \frac{1}{4x^2}$ is 18.
(ii) Using the formula,
[∵ (x - y)3 = x3 - 3xy(x - y) - y3]
So,
$$\begin{aligned}\Big(2x - \frac{1}{2x}\Big)^3 = (2x)^3 - 3 \times 2x \times \frac{1}{2x}\Big(2x - \frac{1}{2x}\Big) - \Big(\frac{1}{2x}\Big)^3\\[1em] ⇒ \Big(2x - \frac{1}{2x}\Big)^3 = 8x^3 - 3\Big(2x - \frac{1}{2x}\Big) - \frac{1}{8x^3}\end{aligned}$$
Putting $2x - \frac{1}{2x} = 4$
$$\begin{aligned}4^3 = 8x^3 - 3 \times 4 - \frac{1}{8x^3}\\[1em] ⇒ 64 = 8x^3 - 12 - \frac{1}{8x^3}\\[1em] ⇒ 8x^3 - \frac{1}{8x^3} = 64 + 12 \\[1em] ⇒ 8x^3 - \frac{1}{8x^3} = 76\end{aligned}$$
Hence, the value of $8x^3 - \frac{1}{8x^3}$ = 76.
If $3x +\frac{1}{3x} = 3$, find:
(i) $9x^2 +\frac{1}{9x^2}$
(ii) $27x^3 +\frac{1}{27x^3}$
(i) Using the formula,
[∵ (x + y)2 = x2 + 2xy + y2]
So,
$$\begin{aligned}\Big(3x + \frac{1}{3x}\Big)^2 = (3x)^2 + 2 \times 3x \times \frac{1}{3x} + \Big(\frac{1}{3x}\Big)^2\\[1em] ⇒ \Big(3x + \frac{1}{3x}\Big)^2 = 9x^2 + 2 + \frac{1}{9x^2}\end{aligned}$$
Putting the value $3x + \frac{1}{3x} = 3$,we get
$$\begin{aligned}3^2 = 9x^2 + 2 + \frac{1}{9x^2}\\[1em] ⇒ 9 = 9x^2 + 2 + \frac{1}{9x^2}\\[1em] ⇒ 9x^2 + \frac{1}{9x^2} = 9 - 2\\[1em] ⇒ 9x^2 + \frac{1}{9x^2} = 7\end{aligned}$$
Hence, the value of $9x^2 + \frac{1}{9x^2}$ is 7.
(ii) Using the formula,
[∵ (x + y)3 = x3 + 3xy(x + y) + y3]
So,
$$\begin{aligned}\Big(3x + \frac{1}{3x}\Big)^3 = (3x)^3 + 3 \times 3x \times \frac{1}{3x}\Big(3x + \frac{1}{3x}\Big) + \Big(\frac{1}{3x}\Big)^3\\[1em] ⇒ \Big(3x + \frac{1}{3x}\Big)^3 = 27x^3 + 3\Big(3x + \frac{1}{3x}\Big) + \frac{1}{27x^3}\end{aligned}$$
Putting $3x + \frac{1}{3x} = 3$
$$\begin{aligned}3^3 = 27x^3 + 3 \times 3 + \frac{1}{27x^3}\\[1em] ⇒ 27 = 27x^3 + 9 + \frac{1}{27x^3}\\[1em] ⇒ 27x^3 + \frac{1}{27x^3} = 27 - 9 \\[1em] ⇒ 27x^3 + \frac{1}{27x^3} = 18\end{aligned}$$
Hence, the value of $27x^3 + \frac{1}{27x^3}$ is 18.
The sum of the squares of two numbers is 13 and their product is 6. Find :
(i) the sum of the two numbers.
(ii) the difference between them.
(i) Let 2 numbers be x and y. The sum of the squares of two numbers is 13 and their product is 6. So,
x2 + y2 = 13
And,
xy = 6
Using the formula,
[∵ (x + y)2 = x2 + 2xy + y2]
(x + y)2 = (x2 + y2) + 2xy
Putting the value,
⇒ (x + y)2 = 13 + 2 $\times$ 6
⇒ (x + y)2 = 13 + 12
⇒ (x + y)2 = 25
⇒ x + y = $\sqrt{25}$
⇒ x + y = 5 or -5
Hence, the sum of the two numbers = 5 or -5.
(ii) Using the formula,
[∵ (x - y)2 = x2 - 2xy + y2]
(x - y)2 = (x2 + y2) - 2xy
Putting the value,
⇒ (x - y)2 = 13 - 2 $\times$ 6
⇒ (x - y)2 = 13 - 12
⇒ (x - y)2 = 1
⇒ x - y = $\sqrt{1}$
⇒ x - y = 1 or -1
Hence, the difference of the two numbers = 1 or -1.
If a is positive and $a^2 +\frac{1}{a^2}= 18$; then the value of $a -\frac{1}{a}$ is:
Using the formula,
[∵(x - y)2 = x2 - 2xy + y2]
$$\begin{aligned}\Big(a -\frac{1}{a}\Big)^2 = a^2 - 2 \times a \times \frac{1}{a} + \Big(\frac{1}{a}\Big)^2\\[1em] = a^2 - \frac{2a}{a} + \Big(\frac{1}{a}\Big)^2\\[1em] = a^2 - 2 + \Big(\frac{1}{a}\Big)^2\\[1em]\end{aligned}$$
Putting the value, $a^2 +\frac{1}{a^2}= 18$
$$\begin{aligned}\Rightarrow \Big(a -\frac{1}{a}\Big)^2 = a^2 + \Big(\frac{1}{a}\Big)^2 - 2\\[1em] \Rightarrow \Big(a -\frac{1}{a}\Big)^2 = 18 - 2\\[1em] \Rightarrow \Big(a -\frac{1}{a}\Big)^2 = 16\\[1em] \Rightarrow \Big(a -\frac{1}{a}\Big) = \sqrt{16}\\[1em] \Rightarrow \Big(a -\frac{1}{a}\Big) = 4 \text{ or} -4\end{aligned}$$
From options, only value matching is 4.
So, $\Big(a -\frac{1}{a}\Big) = 4$
Hence, option 1 is the correct option.
(x + y)(x - y)(x2 + y2)(x4 + y4) is equal to:
Using the formula,
[∵(x - y)(x + y) = x2 - y2]
(x + y)(x - y)(x2 + y2)(x4 + y4)
⇒ [(x + y)(x - y)](x2 + y2)(x4 + y4)
⇒ [(x2 - y2)(x2 + y2)](x4 + y4)
⇒ [(x4 - y4)(x4 + y4)]
⇒ (x8 - y8)
Hence, option 3 is the correct option.
The value of 102 x 98 is:
Using the formula,
[∵(x - y)(x + y) = x2 - y2]
102 x 98 = (100 + 2)(100 - 2)
= 1002 - 22
= 10000 - 4
= 9996
Hence, option 4 is the correct option.
(x + 3) (x + 3) - (x - 2) (x - 2) is equal to:
Using the formula,
[∵(x + y) = x2 + y2 + 2xy]
And,
[∵(x - y) = x2 + y2 - 2xy]
(x + 3) (x + 3) - (x - 2) (x - 2)
⇒ [(x + 3)2] - [(x - 2)2]
⇒ [x2 + 32 + 2 $\times$ x $\times$ 3] - [x2 + 22 - 2 $\times$ x $\times$ 2]
⇒ [x2 + 9 + 6x] - [x2 + 4 - 4x]
⇒ x2 + 9 + 6x - x2 - 4 + 4x
⇒ (x2 - x2) + (9 - 4) + (6x + 4x)
⇒ 5 + 10x
⇒ 10x + 5
Hence, option 1 is the correct option.
If 5a = 302 - 252, the value of a is:
Using the formula,
[∵(x - y)(x + y) = x2 - y2]
5a = 302 - 252
⇒ 5a = (30 - 25) ( 30 + 25)
⇒ 5a = 5 x 55
⇒ 5a = 275
⇒ a = $\frac{275}{5}$
⇒ a = 55
Hence, option 3 is the correct option.
Statement 1: Cube of a binomial : (a - b)3 = a3 + 3a2b - 3ab2 - b3
Statement 2: (a - b)2 - (a + b)2 = 4ab.
Which of the following options is correct?
Cube of a binomial:
(a - b)3 = a3 - b3 - 3ab(a - b)
= a3 - b3 - 3a2b + 3ab2
= a3 - 3a2b + 3ab2 - b3
So, statement 1 is false.
Solving,
(a - b)2 - (a + b)2
= [a2 - 2ab + b2] - [a2 + 2ab + b2]
= a2 - 2ab + b2 - a2 - 2ab - b2
= -4ab.
So, statement 2 is false.
Hence, Option 2 is the correct option.
Assertion (A) : Using appropriate identity, we get 22.5 x 21.5 = 484.75.
Reason (R) : The product: (x + y)(x - y) = x2 - y2.
Solving,
(x + y)(x - y) = x(x - y) + y(x - y)
= x2 - xy + xy - y2
= x2 - y2
So, reason (R) is true.
Solving,
⇒ 22.5 x 21.5
⇒ (22 + 0.5)(22 - 0.5)
Using the identity; (x + y)(x - y) = x2 - y2
= (22)2 - (0.5)2
= 484 - 0.25
= 483.75
So, assertion (A) is false.
Hence, option 4 is the correct option.
Assertion (A) : If we add 9 with 49x2 - 42x, the resultant expression will be a perfect square expression.
Reason (R) : The product of the sum and difference of the same two terms = Difference of their squares.
Given: 49x2 - 42x
Adding 9 in it the term the expression becomes
⇒ 49x2 - 42x + 9
⇒ (7x)2 - 2 × (7x) × 3 + (3)2
⇒ (7x - 3)2
The resultant expression will be a perfect square expression.
So, assertion (A) is true.
The product of the sum and difference of the same two terms = Difference of their squares.
(a + b)(a - b) = a2 - b2
So, reason (R) is true but reason (R) is not the correct explanation of assertion (A).
Hence, option 2 is the correct option.
Assertion (A) : If the volume of a cube is a3 + b3 + 3ab(a + b), then the edge of the cube is (a + b)
Reason (R) : (1st term + 2nd term)3 = (1st term)3 + 3(1st term)2 .(2nd term) + 3(2nd term)2 .(1st term) + (2nd term)3.
Volume of a cube = (edge)3
⇒ (edge)3 = a3 + b3 + 3ab(a + b)
⇒ (edge)3 = (a + b)3
⇒ edge = $\sqrt[3]{(a + b)^3}$
⇒ edge = (a + b)
So, assertion (A) is true.
Using identity,
⇒ (a + b)3 = a3 + b3 + 3ab(a + b)
⇒ (a + b)3 = a3 + 3a2b + 3ab2 + b3
Substitute a = 1st term and b = 2nd term
(1st term + 2nd term)3 = (1st term)3 + 3(1st term)2 .(2nd term) + 3(2nd term)2 .(1st term) + (2nd term)3
So, reason (R) is true and, reason (R) is the correct explanation of assertion (A).
Hence, option 1 is the correct option.
Assertion (A) : 687 x 687 - 313 x 313 = 37400
Reason (R) : The product of the sum and difference of the same two terms = Difference of their squares.
We know that,
The product of the sum and difference of the same two terms = Difference of their squares.
(a + b)(a - b) = a2 - b2
So, reason (R) is true.
Given,
⇒ 687 x 687 - 313 x 313 = 34700
Solving, L.H.S.
⇒ 687 x 687 - 313 x 313
⇒ 6872 - 3132
⇒ (687 - 313)(687 + 313)
⇒ 374 x 1000
⇒ 374000.
Since, L.H.S. ≠ R.H.S.
So, assertion (A) is false.
Hence, option 4 is the correct option.
Evaluate :
$\Big(3x +\frac{1}{2}\Big)$ $\Big(2x +\frac{1}{3}\Big)$
$$\begin{aligned}\Big(3x +\frac{1}{2}\Big) \Big(2x +\frac{1}{3}\Big) \\[1em] = 3x \Big(2x +\frac{1}{3}\Big) + \frac{1}{2} \Big(2x +\frac{1}{3}\Big) \\[1em] = 3x \times 2x + 3x \times \frac{1}{3} + \frac{1}{2} \times 2x + \frac{1}{2} \times \frac{1}{3} \\[1em] = 6x^2 + \frac{3x}{3} + \frac{2x}{2} + \frac{1 \times 1}{2 \times 3} \\[1em] = 6x^2 + x + x + \frac{1}{6} \\[1em] = 6x^2 + 2x + \frac{1}{6}\end{aligned}$$
Hence, $\Big(3x +\frac{1}{2}\Big) \Big(2x +\frac{1}{3}\Big) = 6x^2 + 2x + \frac{1}{6}$
Evaluate :
(2a + 0.5)(7a - 0.3)
(2a + 0.5)(7a - 0.3)
= 2a (7a - 0.3) + 0.5 (7a - 0.3)
= 14a(1+1) - 0.6a + 3.5a - 0.15
= 14a2 + 2.9a - 0.15
Hence, (2a + 0.5)(7a - 0.3) = 14a2 + 2.9a - 0.15
Evaluate :
(9 - y) (7 + y)
(9 - y) (7 + y)
= 9 (7 + y) - y (7 + y)
= 63 + 9y - 7y - y(1+1)
= 63 + 2y - y2
Hence,(9 - y) (7 + y) = 63 + 2y - y2
Evaluate :
(2 - z) (15 - z)
(2 - z) (15 - z)
= 2 (15 - z) - z (15 - z)
= 30 - 2z - 15z + z(1+1)
= 30 - 17z + z2
Hence, (2 - z) (15 - z) = 30 - 17z + z2
Evaluate :
(a2 + 5) (a2 - 3)
(a2 + 5) (a2 - 3)
= a2 (a2 - 3) + 5 (a2 - 3)
= a(2+2) - 3a2 + 5a2 - 15
= a4 + 2a2 - 15
Hence, (a2 + 5) (a2 - 3) = a4 + 2a2 - 15
Evaluate :
(4 - ab) (8 + ab)
(4 - ab) (8 + ab)
= 4 (8 + ab) - ab (8 + ab)
= 32 + 4ab - 8ab - a(1+1)b(1+1)
= 32 - 4ab - a2b2
Hence, (4 - ab) (8 + ab) = 32 - 4ab - a2b2
Evaluate :
(5xy - 7) (7xy + 9)
(5xy - 7) (7xy + 9)
= 5xy (7xy + 9) - 7 (7xy + 9)
= 35x(1+1)y(1+1) + 45xy - 49xy - 63
= 35x2y2 - 4xy - 63
Hence, (5xy - 7) (7xy + 9) = 35x2y2 - 4xy - 63
Evaluate :
(3a2 - 4b2) (8a2 - 3b2)
(3a2 - 4b2) (8a2 - 3b2)
= 3a2 (8a2 - 3b2) - 4b2 (8a2 - 3b2)
= 24a(2+2) - 9a2b2 - 32a2b2 + 12b(2+2)
= 24a4 - 41a2b2 + 12b4
Hence, (3a2 - 4b2) (8a2 - 3b2) = 24a4 - 41a2b2 + 12b4
Find the square of:
$3x + \frac{2}{y}$
Using the formula,
[∵ (x + y)2 = x2 + 2xy + y2]
$$\begin{aligned}\Big(3x +\frac{2}{y}\Big)^2\\[1em] = (3x)^2 + 2 \times 3x \times \frac{2}{y} + \Big(\frac{2}{y}\Big)^2\\[1em] = 9x^2 + \frac{2 \times 6x}{y} + \frac{4}{y^2}\\[1em] = 9x^2 + \frac{12x}{y} + \frac{4}{y^2}\end{aligned}$$
Hence, $\Big(3x +\frac{2}{y}\Big)^2= 9x^2 + \frac{12x}{y} + \frac{4}{y^2}$
Find the square of:
$\frac{5a}{6b} -\frac{6b}{5a}$
Using the formula,
[∵ (x - y)2 = x2 - 2xy + y2]
$$\begin{aligned}\Big(\frac{5a}{6b} - \frac{6b}{5a}\Big)^2\\[1em] = \Big(\frac{5a}{6b}\Big)^2 - 2 \times \frac{5a}{6b} \times \frac{6b}{5a} + \Big(\frac{6b}{5a}\Big)^2\\[1em] = \frac{25a^2}{36b^2} - \frac{2 \times 5a \times 6b}{6b \times 5a} + \frac{36b^2}{25a^2}\\[1em] = \frac{25a^2}{36b^2} - \frac{60ab}{30ab} + \frac{36b^2}{25a^2}\\[1em] = \frac{25a^2}{36b^2} - 2 + \frac{36b^2}{25a^2}\end{aligned}$$
Hence, $\Big(\frac{5a}{6b} - \frac{6b}{5a}\Big)^2 = \frac{25a^2}{36b^2} - 2 + \frac{36b^2}{25a^2}$
Find the square of:
$2m^2 -\frac{2}{3}n^2$
Using the formula,
[∵ (x - y)2 = x2 - 2xy + y2]
$$\begin{aligned}\Big(2m^2 -\frac{2}{3}n^2\Big)^2\\[1em] = (2m^2)^2 - 2 \times 2m^2 \times \frac{2}{3}n^2 + \Big(\frac{2}{3}n^2\Big)^2\\[1em] = 4m^4 - \frac{2 \times 2m^2 \times 2n^2}{3} + \frac{4}{9}n^4\\[1em] = 4m^4 - \frac{8}{3}m^2n^2 + \frac{4}{9}n^4\end{aligned}$$
Hence, $\Big(2m^2 -\frac{2}{3}n^2\Big)^2= 4m^4 - \frac{8}{3}m^2n^2 + \frac{4}{9}n^4$
Find the square of:
$5x +\frac{1}{5x}$
Using the formula,
[∵ (x + y)2 = x2 + 2xy + y2]
$$\begin{aligned}\Big(5x +\frac{1}{5x}\Big)^2\\[1em] = (5x)^2 + 2 \times 5x \times \frac{1}{5x} + \Big(\frac{1}{5x}\Big)^2\\[1em] = 25x^2 + \frac{10x}{5x} + \frac{1}{25x^2}\\[1em] = 25x^2 + 2 + \frac{1}{25x^2}\end{aligned}$$
Hence, $\Big(5x +\frac{1}{5x}\Big)^2= 25x^2 + 2 + \frac{1}{25x^2}$
Find the square of:
$8x +\frac{3}{2}y$
Using the formula,
[∵ (x + y)2 = x2 + 2xy + y2]
$$\begin{aligned}\Big(8x +\frac{3}{2}y\Big)^2\\[1em] = (8x)^2 + 2 \times 8x \times \frac{3}{2}y + \Big(\frac{3}{2}y\Big)^2\\[1em] = 64x^2 + \frac{48xy}{2} + \frac{9}{4}y^2\\[1em] = 64x^2 + 24xy + \frac{9}{4}y^2\end{aligned}$$
Hence, $\Big(8x +\frac{3}{2}y\Big)^2= 64x^2 + 24xy + \frac{9}{4}y^2$
Find the square of:
607
Using the formula,
[∵ (x + y)2 = x2 + 2xy + y2]
6072
= (600 + 7)2
= (600)2 + 2 x 600 x 7 + (7)2
= 3,60,000 + 8,400 + 49
= 3,68,400 + 49
= 3,68,449
Hence, 6072 = 3,68,449
Find the square of:
391
Using the formula,
[∵ (x - y)2 = x2 - 2xy + y2]
3912
= (400 - 9)2
= (400)2 - 2 x 400 x 9 + (9)2
= 1,60,000 - 7,200 + 81
= 1,52,800 + 81
=1,52,881
Hence, 3912 = 1,52,881
Find the square of:
9.7
Using the formula,
[∵ (x - y)2 = x2 - 2xy + y2]
9.72
= (10 - 0.3)2
= (10)2 - 2 x 10 x 0.3 + (0.3)2
= 100 - 6 + 0.09
= 94 + 0.09
= 94.09
Hence, 9.72 = 94.09
If $a +\frac{1}{a}= 2$, find :
(i) $a^2 +\frac{1}{a^2}$
(ii) $a^4 +\frac{1}{a^4}$
(i) Using the formula,
[∵ (x + y)2 = x2 + 2xy + y2]
So,
$$\begin{aligned}\Big(a + \frac{1}{a}\Big)^2 = a^2 + 2 \times a \times \frac{1}{a} + \frac{1}{a^2}\\[1em] ⇒ \Big(a + \frac{1}{a}\Big)^2 = a^2 + 2 + \frac{1}{a^2}\end{aligned}$$
Putting the value $a + \frac{1}{a} = 2$,we get
$$\begin{aligned}2^2 = a^2 + 2 + \frac{1}{a^2}\\[1em] ⇒ 4 = a^2 + 2 + \frac{1}{a^2}\\[1em] ⇒ a^2 + \frac{1}{a^2} = 4 - 2 \\[1em] ⇒ a^2 + \frac{1}{a^2} = 2\end{aligned}$$
Hence, the value of $a^2 + \frac{1}{a^2}$ = 2.
(ii) Using the formula,
[∵ (x + y)2 = x2 + 2xy + y2]
So,
$$\begin{aligned}\Big(a^2 + \frac{1}{a^2}\Big)^2 = (a^2)^2 + 2 \times a^2 \times \frac{1}{a^2} + \Big(\frac{1}{a^2}\Big)^2\\[1em] ⇒ \Big(a^2 + \frac{1}{a^2}\Big)^2 = a^4 + 2 + \frac{1}{a^4}\end{aligned}$$
Putting the value $a^2 + \frac{1}{a^2} = 2$,we get
$$\begin{aligned}2^2 = a^4 + 2 + \frac{1}{a^4}\\[1em] ⇒ 4 = a^4 + 2 + \frac{1}{a^4}\\[1em] ⇒ a^4 + \frac{1}{a^4} = 4 - 2 \\[1em] ⇒ a^4 + \frac{1}{a^4} = 2\end{aligned}$$
Hence, the value of $a^4 + \frac{1}{a^4}$ = 2.
If $m -\frac{1}{m}= 5$, find:
(i) $m^2 +\frac{1}{m^2}$
(ii) $m^4 +\frac{1}{m^4}$
(iii) $m^2 -\frac{1}{m^2}$
(i) Using the formula,
[∵ (x - y)2 = x2 - 2xy + y2]
So,
$$\begin{aligned}\Big(m - \frac{1}{m}\Big)^2 = m^2 - 2 \times m \times \frac{1}{m} + \frac{1}{m^2}\\[1em] ⇒ \Big(m - \frac{1}{m}\Big)^2 = m^2 - 2 + \frac{1}{m^2}\end{aligned}$$
Putting the value $m - \frac{1}{m} = 5$,we get
$$\begin{aligned}5^2 = m^2 - 2 + \frac{1}{m^2}\\[1em] ⇒ 25 = m^2 - 2 + \frac{1}{m^2}\\[1em] ⇒ m^2 + \frac{1}{m^2} = 25 + 2 \\[1em] ⇒ m^2 + \frac{1}{m^2} = 27\end{aligned}$$
Hence, the value of $m^2 + \frac{1}{m^2}$ = 27.
(ii) Using the formula,
[∵ (x + y)2 = x2 + 2xy + y2]
So,
$$\begin{aligned}\Big(m^2 + \frac{1}{m^2}\Big)^2 = (m^2)^2 + 2 \times m^2 \times \frac{1}{m^2} + \Big(\frac{1}{m^2}\Big)^2\\[1em] ⇒ \Big(m^2 + \frac{1}{m^2}\Big)^2 = m^4 + 2 + \frac{1}{m^4}\end{aligned}$$
Putting the value $m^2 + \frac{1}{m^2} = 27$,we get
$$\begin{aligned}27^2 = m^4 + 2 + \frac{1}{m^4}\\[1em] ⇒ 729 = m^4 + 2 + \frac{1}{m^4}\\[1em] ⇒ m^4 + \frac{1}{m^4} = 729 - 2 \\[1em] ⇒ m^4 + \frac{1}{m^4} = 727\end{aligned}$$
Hence, the value of $m^4 + \frac{1}{m^4}$ = 727.
(iii) Using the formula,
[∵ (x + y)2 = x2 + 2xy + y2]
So,
$$\begin{aligned}\Big(m + \frac{1}{m}\Big)^2 = m^2 + 2 \times m \times \frac{1}{m} + \frac{1}{m}^2\\[1em] ⇒ \Big(m + \frac{1}{m}\Big)^2 = m^2 + 2 + \frac{1}{m^2}\\[1em] ⇒ \Big(m + \frac{1}{m}\Big)^2 = m^2 + \frac{1}{m^2} + 2\end{aligned}$$
Putting the value $m^2 + \frac{1}{m^2} = 27$, we get
$$\begin{aligned}⇒ \Big(m + \frac{1}{m}\Big)^2 = 27 + 2 \\[1em] ⇒ \Big(m + \frac{1}{m}\Big)^2 = 29 \\[1em] ⇒ \Big(m + \frac{1}{m}\Big) = \sqrt{29} \\[1em]\end{aligned}$$
Now, using the formula,
[∵ (x2 - y2) = (x - y)(x + y)]
So,
$\Big(m^2 - \frac{1}{m^2}\Big) = \Big(m - \frac{1}{m}\Big)\Big(m + \frac{1}{m}\Big)$
Putting the value, $\Big(m - \frac{1}{m}\Big) = 5$ and $\Big(m + \frac{1}{m}\Big) = \sqrt{29}$
$$\begin{aligned}\Big(m^2 - \frac{1}{m^2}\Big) = (5) \times (\sqrt{29})\\[1em] m^2 - \frac{1}{m^2} = 5\sqrt{29}\end{aligned}$$
Hence, the value of $m^2 - \frac{1}{m^2} = 5\sqrt{29}$.
If a2 + b2 = 41 and ab = 4, find:
(i) a - b
(ii) a + b
(i) Using the formula,
[∵(x - y)2 = x2 - 2xy + y2]
So,
(a - b)2 = a2 - 2ab + b2
Putting the value, a2 + b2 = 41 and ab = 4
⇒ (a - b)2 = (a2 + b2) - 2ab
⇒ (a - b)2 = (41) - 2 $\times$ 4
⇒ (a - b)2 = 41 - 8
⇒ (a - b)2 = 33
⇒ a - b = $\sqrt{33}$
⇒ a - b = $\sqrt{33}$
Hence, the value of a - b = $\sqrt{33}$.
(ii) Using the formula,
[∵(x + y)2 = x2 + 2xy + y2]
So,
(a + b)2 = a2 + 2ab + b2
Putting the value, a2 + b2 = 41 and ab = 4
⇒ (a + b)2 = (a2 + b2) + 2ab
⇒ (a + b)2 = (41) + 2 $\times$ 4
⇒ (a + b)2 = 41 + 8
⇒ (a + b)2 = 49
⇒ a + b = $\sqrt{49}$
⇒ a + b = 7
Hence, the value of a + b = 7.
If $2a +\frac{1}{2a}= 8$, find:
(i) $4a^2 +\frac{1}{4a^2}$
(ii) $16a^4 +\frac{1}{16a^4}$
(i) Using the formula,
[∵ (x + y)2 = x2 + 2xy + y2]
So,
$$\begin{aligned}\Big(2a + \frac{1}{2a}\Big)^2 = (2a)^2 + 2 \times 2a \times \frac{1}{2a} + \Big(\frac{1}{2a}\Big)^2\\[1em] ⇒ \Big(2a + \frac{1}{2a}\Big)^2 = 4a^2 + 2 + \frac{1}{4a^2}\end{aligned}$$
Putting the value $2a + \frac{1}{2a} = 8$, we get
$$\begin{aligned}8^2 = 4a^2 + 2 + \frac{1}{4a^2}\\[1em] ⇒ 64 = 4a^2 + 2 + \frac{1}{4a^2}\\[1em] ⇒ 4a^2 + \frac{1}{4a^2} = 64 - 2 \\[1em] ⇒ 4a^2 + \frac{1}{4a^2} = 62\end{aligned}$$
Hence, the value of $4a^2 + \frac{1}{4a^2}$ is 62.
(ii) Using the formula,
[∵ (x + y)2 = x2 + 2xy + y2]
So,
$$\begin{aligned}\Big(4a^2 + \frac{1}{4a^2}\Big)^2 = (4a^2)^2 + 2 \times 4a^2 \times \frac{1}{4a^2} + \Big(\frac{1}{4a^2}\Big)^2\\[1em] ⇒ \Big(4a^2 + \frac{1}{4a^2}\Big)^2 = 16a^4 + 2 + \frac{1}{16a^4}\end{aligned}$$
Putting the value $4a^2 + \frac{1}{4a^2} = 62$, we get
$$\begin{aligned}62^2 = 16a^4 + 2 + \frac{1}{16a^4}\\[1em] ⇒ 3,844 = 16a^4 + 2 + \frac{1}{16a^4}\\[1em] ⇒ 16a^4 + \frac{1}{16a^4} = 3,844 - 2 \\[1em] ⇒ 16a^4 + \frac{1}{16a^4} = 3,842\end{aligned}$$
Hence, the value of $16a^4 + \frac{1}{16a^4}$ is 3,842.
If $3x -\frac{1}{3x}= 5$, find:
(i) $9x^2 +\frac{1}{9x^2}$
(ii) $81x^4 +\frac{1}{81x^4}$
(i) Using the formula,
[∵ (x - y)2 = x2 - 2xy + y2]
So,
$$\begin{aligned}\Big(3x - \frac{1}{3x}\Big)^2 = (3x)^2 - 2 \times 3x \times \frac{1}{3x} + \Big(\frac{1}{3x}\Big)^2\\[1em] ⇒ \Big(3x - \frac{1}{3x}\Big)^2 = 9x^2 - 2 + \frac{1}{9x^2}\end{aligned}$$
Putting the value $3x - \frac{1}{3x} = 5$,we get
$$\begin{aligned}5^2 = 9x^2 - 2 + \frac{1}{9x^2}\\[1em] ⇒ 25 = 9x^2 - 2 + \frac{1}{9x^2}\\[1em] ⇒ 9x^2 + \frac{1}{9x^2} = 25 + 2 \\[1em] ⇒ 9x^2 + \frac{1}{9x^2} = 27\end{aligned}$$
Hence, the value of $9x^2 + \frac{1}{9x^2}$ is 27.
(ii) Using the formula,
[∵ (x + y)2 = x2 + 2xy + y2]
So,
$$\begin{aligned}\Big(9x^2 + \frac{1}{9x^2}\Big)^2 = (9x^2)^2 + 2 \times 9x^2 \times \frac{1}{9x^2} + \Big(\frac{1}{9x^2}\Big)^2\\[1em] ⇒ \Big(9x^2 + \frac{1}{9x^2}\Big)^2 = 81x^4 + 2 + \frac{1}{81x^4}\end{aligned}$$
Putting the value $9x^2 + \frac{1}{9x^2} = 27$,we get
$$\begin{aligned}27^2 = 81x^4 + 2 + \frac{1}{81x^4}\\[1em] ⇒ 729 = 81x^4 + 2 + \frac{1}{81x^4}\\[1em] ⇒ 81x^4 + \frac{1}{81x^4} = 729 - 2 \\[1em] ⇒ 81x^4 + \frac{1}{81x^4} = 727\end{aligned}$$
Hence, the value of $81x^4 + \frac{1}{81x^4}$ is 727.
Expand:
(3x - 4y + 5z)2
Using the formula,
[∵ (x - y + z)2 = x2 + y2 + z2 - 2xy - 2yz + 2xz]
⇒ (3x - 4y + 5z)2 = (3x)2 + (4y)2 + (5z)2 - 2 $\times$ (3x) $\times$ (4y) - 2 $\times$ (4y) $\times$ (5z) + 2 $\times$ (5z) $\times$ (3x)
= 9x2 + 16y2 + 25z2 - 24xy - 40yz + 30xz
Hence, (3x - 4y + 5z)2 = 9x2 + 16y2 + 25z2 - 24xy - 40yz + 30xz.
Expand:
(2a - 5b - 4c)2
Using the formula,
[∵ (x - y - z)2 = x2 + y2 + z2 - 2xy + 2yz - 2xz]
⇒ (2a - 5b - 4c)2 = (2a)2 + (5b)2 + (4c)2 - 2 $\times$ (2a) $\times$ (5b) + 2 $\times$ (5b) $\times$ (4c) - 2 $\times$ (4c) $\times$ (2a)
= 4a2 + 25b2 + 16c2 - 20ab + 40bc - 16ca
Hence, (2a - 5b - 4c)2 = 4a2 + 25b2 + 16c2 - 20ab + 40bc - 16ca.
Expand:
(5x + 3y)3
Using the formula,
[∵ (x + y)3 = x3 + y3 + 3x2y + 3xy2]
So,
⇒ (5x + 3y)3 = (5x)3 + (3y)3 + 3 $\times$ (5x)2 $\times$ (3y) + 3 $\times$(5x) $\times$ (3y)2
= 125x3 + 27y3 + 225x2y + 135xy2
Hence, (5x + 3y)3 = 125x3 + 27y3 + 225x2y + 135xy2.
Expand:
(6a - 7b)3
Using the formula,
[∵ (x - y)3 = x3 - y3 - 3x2y + 3xy2]
So,
⇒ (6a - 7b)3 = (6a)3 - (7b)3 - 3 $\times$ (6a)2 $\times$ (7b) + 3 $\times$(6a) $\times$ (7b)2
= 216a3 - 343b3 - 756a2b + 882ab2
Hence, (6a - 7b)3 = 216a3 - 343b3 - 756a2b + 882ab2.
If a + b + c = 9 and ab + bc + ca = 15, find: a2 + b2 + c2.
Using the formula,
[∵(x + y + z)2 = x2 + y2 + z2 + 2xy + 2yz + 2zx]
So,
(a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca
⇒ (a + b + c)2 = a2 + b2 + c2 + 2(ab + bc + ca)
Putting the value a + b + c = 9 and ab + bc + ca = 15, we get
⇒ 92 = a2 + b2 + c2 + 2 x 15
⇒ 81 = a2 + b2 + c2 + 30
⇒ a2 + b2 + c2 = 81 - 30
⇒ a2 + b2 + c2 = 51
Hence, the value of a2 + b2 + c2 is 51.
If a + b + c = 11 and a2 + b2 + c2 = 81, find: ab + bc + ca.
Using the formula,
[∵(x + y + z)2 = x2 + y2 + z2 + 2xy + 2yz + 2zx]
So,
(a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca
⇒ (a + b + c)2 = a2 + b2 + c2 + 2(ab + bc + ca)
Putting the value (a + b + c) = 11 and a2 + b2 + c2 = 81,
⇒ (11)2 = 81 + 2(ab + bc + ca)
⇒ 121 = 81 + 2(ab + bc + ca)
⇒ 2(ab + bc + ca) = 121 - 81
⇒ 2(ab + bc + ca) = 40
⇒ ab + bc + ca = $\frac{40}{2}$
⇒ ab + bc + ca = 20
Hence, the value of (ab + bc + ca) is 20.
If 3x - 4y = 5 and xy = 3, find: 27x3 - 64y3
Using the formula,
[∵ (x - y)3 = x3 - y3 - 3xy(x - y)]
So,
⇒ (3x - 4y)3 = (3x)3 - (4y)3 - 3 $\times$ 3x $\times$ 4y(3x - 4y)
⇒ (3x - 4y)3 = 27x3 - 64y3 - 36xy(3x - 4y)
Putting the value (3x - 4y) = 5 and xy = 3, we get
⇒ (5)3 = 27x3 - 64y3 - 36 $\times$ 3 $\times$ 5
⇒ 125 = 27x3 - 64y3 - 540
⇒ 27x3 - 64y3 = 125 + 540
⇒ 27x3 - 64y3 = 665
Hence, the value of 27x3 - 64y3 is 665.
If a + b = 8 and ab = 15, find: a3 + b3
Using the formula,
[∵ (x + y)3 = x3 + y3 + 3xy(x + y)]
So,
⇒ (a + b)3 = a3 + b3 + 3ab(a + b)
Putting the value (a + b) = 8 and ab = 15, we get
⇒ (8)3 = a3 + b3 + 3 $\times$ 15 $\times$ 8
⇒ 512 = a3 + b3 + 360
⇒ a3 + b3 = 512 - 360
⇒ a3 + b3 = 152
Hence, the value of a3 + b3 is 152.
If 3x + 2y = 9 and xy = 3, find: 27x3 + 8y3
Using the formula,
[∵ (x + y)3 = x3 + y3 + 3xy(x + y)]
So,
⇒ (3x + 2y)3 = (3x)3 + (2y)3 + 3 $\times$ 3x $\times$ 2y(3x + 2y)
⇒ (3x + 2y)3 = 27x3 + 8y3 + 18xy(3x + 2y)
Putting the value (3x + 2y) = 9 and xy = 3, we get
⇒ (9)3 = 27x3 + 8y3 + 18 $\times$ 3 $\times$ 9
⇒ 729 = 27x3 + 8y3 + 486
⇒ 27x3 + 8y3 = 729 - 486
⇒ 27x3 + 8y3 = 243
Hence, the value of 27x3 + 8y3 is 243.
If 5x - 4y = 7 and xy = 8, find: 125x3 - 64y3
Using the formula,
[∵ (x - y)3 = x3 - y3 - 3xy(x - y)]
So,
⇒ (5x - 4y)3 = (5x)3 - (4y)3 - 3 $\times$ 5x $\times$ 4y(5x - 4y)
⇒ (5x - 4y)3 = 125x3 - 64y3 - 60xy(5x - 4y)
Putting the value (5x - 4y) = 7 and xy = 8, we get
⇒ (7)3 = 125x3 - 64y3 - 60 $\times$ 8 $\times$ 7
⇒ 343 = 125x3 - 64y3 - 3,360
⇒ 125x3 - 64y3 = 343 + 3,360
⇒ 125x3 - 64y3 = 3,703
Hence, the value of 125x3 - 64y3 is 3,703.
The difference between two numbers is 5 and their product is 14. Find the difference between their cubes.
Let the two numbers be x and y.
So,
x - y = 5
And,
xy = 14
Using the formula,
[∵ (x - y)3 = x3 - y3 - 3xy(x - y)]
So,
⇒ (x - y)3 = x3 - y3 - 3xy(x - y)
Putting the value (x - y) = 5 and xy = 14, we get
⇒ (5)3 = x3 - y3 - 3 $\times$ 14 $\times$ 5
⇒ 125 = x3 - y3 - 210
⇒ x3 - y3 = 125 + 210
⇒ x3 - y3 = 335
Hence, the difference between their cubes is 335.
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