Expansions — Exercise 3(B)

Expand:

(i) (3a + 5b)³

(ii) (2p - 3q)³

(iii) $\Big(2x + \dfrac{1}{3x}\Big)^3$

(iv) (3ab - 2c)³

(v) $\Big(3a - \dfrac{1}{a}\Big)^3$

(vi) $\Big(\dfrac{1}{2}x - \dfrac{2}{3}y\Big)^3$

Answer

(i) Given,

(3a + 5b)³

Using identity :

(a + b)³ = a³ + b³ + 3ab(a + b)

⇒ (3a + 5b)³ = (3a)³ + (5b)³ + 3 × 3a × 5b × (3a + 5b)

⇒ (3a + 5b)³ = 27a³ + 125b³ + 45ab × (3a + 5b)

⇒ (3a + 5b)³ = 27a³ + 135a² b + 225ab² + 125b³

Hence, (3a + 5b)³ = 27a³ + 135a²b + 225ab² + 125b³.

(ii) Given,

(2p - 3q)³

Using identity :

(a - b)³ = a³ - b³ - 3a²b + 3ab²

⇒ (2p - 3q)³ = [(2p)³ - (3q)³ - 3 × (2p)² (3q) + 3 × (2p) × (3q)²]

⇒ (2p - 3q)³ = 8p³ - 27q³ - 3 × 4p² (3q) + 3 × (2p) × 9q²

⇒ (2p - 3q)³ = 8p³ - 36p²q + 54pq² - 27q³

Hence, (2p - 3q)³ = 8p³ - 36p²q + 54pq² - 27q³.

(iii) Given,

$\Big(2x + \dfrac{1}{3x}\Big)^3$

Using identity :

(a + b)³ = a³ + b³ + 3a²b + 3ab²

$\Rightarrow \Big(2x + \dfrac{1}{3x}\Big)^3 = \Big[(2x)^3 + \Big(\dfrac{1}{3x}\Big)^3 + 3 \times (2x)^2 \times \Big(\dfrac{1}{3x}\Big) + 3 \times 2x \times \Big(\dfrac{1}{3x}\Big)^2 \Big]\\[1em] \Rightarrow \Big(2x + \dfrac{1}{3x}\Big)^3 = 8x^3 + \dfrac{1}{27x^3} + 3 \times 4x^2 \times \Big(\dfrac{1}{3x}\Big) + 3 \times 2x \times \Big(\dfrac{1}{9x^2}\Big) \\[1em] \Rightarrow \Big(2x + \dfrac{1}{3x}\Big)^3 = 8x^3 + 4x + \dfrac{2}{3x} + \dfrac{1}{27x^3} \\[1em]$

Hence, $\Big(2x + \dfrac{1}{3x}\Big)^3 = 8x^3 + 4x + \dfrac{2}{3x} + \dfrac{1}{27x^3}$.

(iv) Given,

(3ab - 2c)³

Using identity :

(a - b)³ = a³ - b³ - 3a²b + 3ab²

⇒ (3ab - 2c)³ = [(3ab)³ - (2c)³ - 3 × (3ab)² (2c) + 3 × (3ab) × (2c)²]

⇒ (3ab - 2c)³ = 27a³b³ - 8c³ - (9a²b²) × 6c + 3 × (3ab) × 4c²

⇒ (3ab - 2c)³ = 27a³b³ - 54a²b²c + 36 abc² - 8c³

Hence, (3ab - 2c)³ = 27a³b³ - 54a²b²c + 36 abc² - 8c³.

(v) Given,

$\Big(3a - \dfrac{1}{a}\Big)^3$

Using identity :

(a - b)³ = a³ - b³ - 3a²b + 3ab²

$\Rightarrow \Big(3a - \dfrac{1}{a}\Big)^3 = \Big[(3a)^3 - \Big(\dfrac{1}{a}\Big)^3 - 3 \times (3a)^2 \times \Big(\dfrac{1}{a}\Big) + 3 \times 3a \times \Big(\dfrac{1}{a}\Big)^2 \Big]\\[1em] \Rightarrow \Big(3a - \dfrac{1}{a}\Big)^3 = 27a^3 - \dfrac{1}{a^3} - 3 \times 9a^2 \times \Big(\dfrac{1}{a}\Big) + 3 \times 3a \times \Big(\dfrac{1}{a^2}\Big) \\[1em] \Rightarrow \Big(3a - \dfrac{1}{a}\Big)^3 = 27a^3 - 27a + \dfrac{9}{a} - \dfrac{1}{a^3} \\[1em]$

Hence, $\Big(3a - \dfrac{1}{a}\Big)^3 = 27a^3 - 27a + \dfrac{9}{a} - \dfrac{1}{a^3}$.

(vi) Given,

$\Big(\dfrac{1}{2}x - \dfrac{2}{3}y\Big)^3$

Using identity :

(a - b)³ = a³ - b³ - 3a²b + 3ab²

$\Rightarrow \Big(\dfrac{1}{2}x - \dfrac{2}{3}y\Big)^3 = \Big[\Big(\dfrac{1}{2}x\Big)^3 - \Big(\dfrac{2}{3}y\Big)^3 - 3 \times \Big(\dfrac{1}{2}x\Big)^2 \times \Big(\dfrac{2}{3}y\Big) + 3 \times \Big(\dfrac{1}{2}x\Big) \times \Big(\dfrac{2}{3}y\Big)^2 \Big]\\[1em] \Rightarrow \Big(\dfrac{1}{2}x - \dfrac{2}{3}y\Big)^3 = \Big(\dfrac{1}{8}x^3\Big) - \Big(\dfrac{8}{27}y^3\Big) - 3 \times \Big(\dfrac{1}{4}x^2\Big) \times \Big(\dfrac{2}{3}y\Big) + 3 \times \Big(\dfrac{1}{2}x\Big) \times \Big(\dfrac{4}{9}y^2\Big) \\[1em] \Rightarrow \Big(\dfrac{1}{2}x - \dfrac{2}{3}y\Big)^3 = \dfrac{1}{8}x^3 - \dfrac{8}{27}y^3 - \dfrac{6}{12}x^2y + \dfrac{12}{18}xy^2 \\[1em] \Rightarrow \Big(\dfrac{1}{2}x - \dfrac{2}{3}y\Big)^3 = \dfrac{1}{8}x^3 - \dfrac{8}{27}y^3 - \dfrac{1}{2}x^2y + \dfrac{2}{3}xy^2 \\[1em]$

Hence, $\Big(\dfrac{1}{2}x - \dfrac{2}{3}y\Big)^3 = \dfrac{1}{8}x^3 - \dfrac{8}{27}y^3 - \dfrac{1}{2}x^2y + \dfrac{2}{3}xy^2$.

If 4a + 3b = 10 and ab = 2, find the value of $64a^3 + 27b^3$.

Answer

Given,

(4a + 3b) = 10

ab = 2

⇒ (4a + 3b)³ = [(4a)³ + (3b)³ + 3 × 4a × 3b × (4a + 3b)]

⇒ (10)³ = 64a³ + 27b³ + 36ab × (10)

⇒ 1000 = 64a³ + 27b³ + 36 × 2 × 10

⇒ 1000 = 64a³ + 27b³ + 720

⇒ 64a³ + 27b³ = 1000 - 720

⇒ 64a³ + 27b³ = 280

Hence, 64a³ + 27b³ = 280.

If 3x – 2y = 5 and xy = 6, find the value of $27x^3 - 8y^3$.

Answer

Given,

3x – 2y = 5

xy = 6

Using identity :

(a - b)³ = a³ - b³ - 3ab(a - b)

⇒ (3x – 2y)³ = [(3x)³ - (2y)³ - 3 × 3x × 2y × (3x - 2y)]

⇒ (5)³ = 27x³ - 8y³ - 18xy × (5)

⇒ 125 = 27x³ - 8y³ - 90xy

⇒ 125 = 27x³ - 8y³ - 90(6)

⇒ 27x³ - 8y³ = 125 + 540

⇒ 27x³ - 8y³ = 665

Hence, 27x³ - 8y³ = 665.

If a + 3b = 6, show that $a^3 + 27b^3 + 54ab = 216$.

Answer

Given,

a + 3b = 6

⇒ (a + 3b)³ = [(a)³ + (3b)³ + 3 × a × 3b × (a + 3b)]

⇒ (6)³ = a³ + 27b³ + 9ab × (a + 3b)

⇒ 216 = a³ + 27b³ + 9ab × (6)

⇒ a³ + 27b³ + 54ab = 216

Hence proved that a³ + 27b³ + 54ab = 216.

If a + 2b + 3c = 0, show that a³ + 8b³ + 27c³ = 18abc.

Answer

We know that,

If x + y + z = 0 then x³ + y³ + z³ = 3xyz ........(1)

Since, (a + 2b + 3c) = 0,

⇒ (a)³ + (2b)³ + (3c)³ = 3 × a × 2b × 3c

⇒ a³ + 8b³ + 27c³ = 18abc.

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

If $x + \dfrac{1}{x} = 3$, find the value of $\Big(x^3 + \dfrac{1}{x^3}\Big)$.

Answer

Given,

$x + \dfrac{1}{x} = 3$

Using identity,

$\Rightarrow \Big(x + \dfrac{1}{x}\Big)^3 = x^3 + \dfrac{1}{x^3} + 3\Big(x + \dfrac{1}{x}\Big) \\[1em] \Rightarrow 3^3 = x^3 + \dfrac{1}{x^3} + 3 \times 3 \\[1em] \Rightarrow 27 = x^3 + \dfrac{1}{x^3} + 9 \\[1em] \Rightarrow x^3 + \dfrac{1}{x^3} = 27 - 9 = 18.$

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

If $x - \dfrac{1}{x} = 5$, find the value of $\Big(x^3 - \dfrac{1}{x^3}\Big)$.

Answer

Given,

$x - \dfrac{1}{x} = 5$

Using identity,

$\Rightarrow \Big(x^3 - \dfrac{1}{x^3}\Big) = \Big(x - \dfrac{1}{x}\Big)^3 + 3\Big(x - \dfrac{1}{x}\Big) \\[1em] \Rightarrow \Big(x^3 - \dfrac{1}{x^3}\Big) = (5)^3 + 3 \times 5 \\[1em] \Rightarrow \Big(x^3 - \dfrac{1}{x^3}\Big) = 125 + 15 \\[1em] \Rightarrow \Big(x^3 - \dfrac{1}{x^3}\Big) = 140$

Hence, $x^3 - \dfrac{1}{x^3} = 140.$

If $x - \dfrac{2}{x} = 6$, find the value of $\Big(x^3 - \dfrac{8}{x^3}\Big)$.

Answer

Given,

$\Rightarrow x - \dfrac{2}{x} = 6$

Upon cubing both sides we get :

$\Rightarrow \Big(x - \dfrac{2}{x}\Big)^3 = 6^3 \\[1em] \Rightarrow (x)^3 - \Big(\dfrac{2}{x}\Big)^3 - 3 \times x \times \dfrac{2}{x} \times \Big(x - \dfrac{2}{x}\Big) = 216 \\[1em] \Rightarrow (x)^3 - \Big(\dfrac{2}{x}\Big)^3 - 6 \times 6 = 216 \\[1em] \Rightarrow x^3 - \dfrac{8}{x^3} - 36 = 216 \\[1em] \Rightarrow x^3 - \dfrac{8}{x^3} = 216 + 36 \\[1em] \Rightarrow x^3 - \dfrac{8}{x^3} = 252.$

Hence, $x^3 - \dfrac{8}{x^3} = 252.$

If $x + \dfrac{1}{x} = 4$, find the values of:

(i) $\Big(x^3 + \dfrac{1}{x^3}\Big)$

(ii) $\Big(x - \dfrac{1}{x}\Big)$

(iii) $\Big(x^3 - \dfrac{1}{x^3}\Big)$

Answer

(i) Given,

$x + \dfrac{1}{x} = 4$

Using identity,

$\Rightarrow \Big(x + \dfrac{1}{x}\Big)^3 = x^3 + \dfrac{1}{x^3} + 3\Big(x + \dfrac{1}{x}\Big) \\[1em] \Rightarrow 4^3 = x^3 + \dfrac{1}{x^3} + 3 \times 4 \\[1em] \Rightarrow 64 = x^3 + \dfrac{1}{x^3} + 12 \\[1em] \Rightarrow x^3 + \dfrac{1}{x^3} = 64 - 12 = 52.$

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

(ii) Given,

$x + \dfrac{1}{x} = 4$

Using identity,

$\Rightarrow \Big(x + \dfrac{1}{x}\Big)^2 - \Big(x - \dfrac{1}{x}\Big)^2 = 4 \\[1em] \Rightarrow (4)^2 - \Big(x - \dfrac{1}{x}\Big)^2 = 4 \\[1em] \Rightarrow 16 - \Big(x - \dfrac{1}{x}\Big)^2 = 4 \\[1em] \Rightarrow \Big(x - \dfrac{1}{x}\Big)^2 = 16 - 4 \\[1em] \Rightarrow \Big(x - \dfrac{1}{x}\Big)^2 = 12 \\[1em] \Rightarrow \Big(x - \dfrac{1}{x}\Big) = \sqrt{12}\\[1em] \Rightarrow \Big(x - \dfrac{1}{x}\Big) = \pm 2\sqrt{3}.$

Hence, $\Big(x - \dfrac{1}{x}\Big) = \pm 2\sqrt{3}.$

(iii) Given,

$\Big(x - \dfrac{1}{x}\Big) = \pm 2\sqrt{3}$

Case 1:

$\Big(x - \dfrac{1}{x}\Big) = 2\sqrt{3}$

We know that,

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

Substituting values we get :

$\Rightarrow \Big(x^3 - \dfrac{1}{x^3}\Big) = (2\sqrt{3})^3 + 3(2\sqrt{3}) \\[1em] \Rightarrow \Big(x^3 - \dfrac{1}{x^3}\Big) = (8 \times 3\sqrt{3}) + (6\sqrt{3}) \\[1em] \Rightarrow \Big(x^3 - \dfrac{1}{x^3}\Big) = (24\sqrt{3}) + (6\sqrt{3}) \\[1em] \Rightarrow \Big(x^3 - \dfrac{1}{x^3}\Big) = 30\sqrt{3} \\[1em]$

Case 1:

$\Big(x - \dfrac{1}{x}\Big) = -2\sqrt{3}$

We know that,

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

Substituting values we get :

$\Rightarrow \Big(x^3 - \dfrac{1}{x^3}\Big) = (-2\sqrt{3})^3 + 3(-2\sqrt{3}) \\[1em] \Rightarrow \Big(x^3 - \dfrac{1}{x^3}\Big) = (-8 \times 3\sqrt{3}) + (-6\sqrt{3}) \\[1em] \Rightarrow \Big(x^3 - \dfrac{1}{x^3}\Big) = (-24\sqrt{3}) - (6\sqrt{3}) \\[1em] \Rightarrow \Big(x^3 - \dfrac{1}{x^3}\Big) = -30\sqrt{3} \\[1em]$

Hence, $\Big(x^3 - \dfrac{1}{x^3}\Big) = \pm 30\sqrt{3}.$

If $a^2 + \dfrac{1}{a^2} = 23$, find the values of:

(i) $\Big(a + \dfrac{1}{a}\Big)$

(ii) $\Big(a^3 + \dfrac{1}{a^3}\Big)$

Answer

(i) Given,

$a^2 + \dfrac{1}{a^2} = 23$

Using identity,

$\Rightarrow \Big(a + \dfrac{1}{a}\Big)^2 = a^2 + \dfrac{1}{a^2} + 2 \\[1em] \Rightarrow \Big(a + \dfrac{1}{a}\Big)^2 = 23 + 2 \\[1em] \Rightarrow \Big(a + \dfrac{1}{a}\Big)^2 = 25 \\[1em] \Rightarrow \Big(a + \dfrac{1}{a}\Big) = \sqrt{25} \\[1em] \Rightarrow \Big(a + \dfrac{1}{a}\Big) = \pm 5$

Hence, $\Big(a + \dfrac{1}{a}\Big) = \pm 5.$

(ii) Given,

$\Big(a + \dfrac{1}{a}\Big) = \pm 5.$

Case 1:

$\Big(a + \dfrac{1}{a}\Big) = +5.$

We know that,

$\Rightarrow \Big(a + \dfrac{1}{a}\Big)^3 = a^3 + \dfrac{1}{a^3} + 3\Big(a + \dfrac{1}{a}\Big)$

Substituting values we get :

$\Rightarrow 5^3 = a^3 + \dfrac{1}{a^3} + 3 \times 5 \\[1em] \Rightarrow 125 = a^3 + \dfrac{1}{a^3} + 15 \\[1em] \Rightarrow a^3 + \dfrac{1}{a^3} = 125 - 15 = 110.$

Case 2:

$\Big(a + \dfrac{1}{a}\Big) = -5.$

We know that,

$\Rightarrow \Big(a + \dfrac{1}{a}\Big)^3 = a^3 + \dfrac{1}{a^3} + 3\Big(a + \dfrac{1}{a}\Big)$

Substituting values we get :

$\Rightarrow (-5)^3 = a^3 + \dfrac{1}{a^3} + 3 \times (-5) \\[1em] \Rightarrow -125 = a^3 + \dfrac{1}{a^3} - 15 \\[1em] \Rightarrow a^3 + \dfrac{1}{a^3} = -125 + 15 = -110.$

Hence, $a^3 + \dfrac{1}{a^3} = \pm 110.$

If $a - \dfrac{1}{a} = \sqrt{5}$, find the values of :

(i) $\Big(a + \dfrac{1}{a}\Big)$

(ii) $\Big(a^3 + \dfrac{1}{a^3}\Big)$

Answer

(i) Given,

$a - \dfrac{1}{a} = \sqrt{5}$

Using identity,

$\Rightarrow \Big(a + \dfrac{1}{a}\Big)^2 - \Big(a - \dfrac{1}{a}\Big)^2 = 4 \\[1em] \Rightarrow \Big(a + \dfrac{1}{a}\Big)^2 - (\sqrt{5})^2 = 4 \\[1em] \Rightarrow \Big(a + \dfrac{1}{a}\Big)^2 - 5 = 4 \\[1em] \Rightarrow \Big(a + \dfrac{1}{a}\Big)^2 = 4 + 5 \\[1em] \Rightarrow \Big(a + \dfrac{1}{a}\Big)^2 = 9 \\[1em] \Rightarrow \Big(a + \dfrac{1}{a}\Big) = \sqrt{9} \\[1em] \Rightarrow \Big(a + \dfrac{1}{a}\Big) = \pm 3$

Hence, $\Big(a + \dfrac{1}{a}\Big) = \pm 3.$

(ii) Given,

$\Big(a + \dfrac{1}{a}\Big) = \pm 3.$

Case 1:

$\Big(a + \dfrac{1}{a}\Big) = +3.$

We know that,

$\Rightarrow \Big(a + \dfrac{1}{a}\Big)^3 = a^3 + \dfrac{1}{a^3} + 3\Big(a + \dfrac{1}{a}\Big)$

Substituting values we get :

$\Rightarrow 3^3 = a^3 + \dfrac{1}{a^3} + 3 \times 3 \\[1em] \Rightarrow 27 = a^3 + \dfrac{1}{a^3} + 9 \\[1em] \Rightarrow a^3 + \dfrac{1}{a^3} = 27 - 9 = 18.$

Case 2:

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

We know that,

$\Rightarrow \Big(a + \dfrac{1}{a}\Big)^3 = a^3 + \dfrac{1}{a^3} + 3\Big(a + \dfrac{1}{a}\Big)$

Substituting values we get :

$\Rightarrow (-3)^3 = a^3 + \dfrac{1}{a^3} + 3 \times (-3) \\[1em] \Rightarrow -27 = a^3 + \dfrac{1}{a^3} - 9 \\[1em] \Rightarrow a^3 + \dfrac{1}{a^3} = -27 + 9 = -18.$

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

If $a^2 + \dfrac{1}{a^2} = 27$, find the values of :

(i) $\Big(a - \dfrac{1}{a}\Big)$

(ii) $\Big(a^3 - \dfrac{1}{a^3}\Big)$

Answer

(i) Given,

$a^2 + \dfrac{1}{a^2} = 27$

Using identity,

$\Rightarrow \Big(a - \dfrac{1}{a}\Big)^2 = a^2 + \dfrac{1}{a^2} - 2 \\[1em] \Rightarrow \Big(a - \dfrac{1}{a}\Big)^2 = 27 - 2 \\[1em] \Rightarrow \Big(a - \dfrac{1}{a}\Big)^2 = 25 \\[1em] \Rightarrow \Big(a - \dfrac{1}{a}\Big)^2 = \sqrt{25} \\[1em] \Rightarrow \Big(a - \dfrac{1}{a}\Big)^2 = \pm 5$

Hence, $\Big(a - \dfrac{1}{a}\Big)^2 = \pm 5.$

(ii) Given,

$\Big(a + \dfrac{1}{a}\Big) = \pm 5.$

Case 1:

$\Big(a + \dfrac{1}{a}\Big) = +5.$

We know that,

$\Rightarrow \Big(a^3 - \dfrac{1}{a^3}\Big) = \Big(a - \dfrac{1}{a}\Big)^3 + 3 \times a \times \dfrac{1}{a} \Big(a - \dfrac{1}{a}\Big)$

Substituting values we get :

$\Rightarrow \Big(a^3 - \dfrac{1}{a^3}\Big) = (5)^3 + 3(5) \\[1em] \Rightarrow \Big(a^3 - \dfrac{1}{a^3}\Big) = (125) + (15) \\[1em] \Rightarrow \Big(a^3 - \dfrac{1}{a^3}\Big) = 140.$

Case 2:

$\Big(a + \dfrac{1}{a}\Big) = -5.$

We know that,

$\Rightarrow \Big(a^3 - \dfrac{1}{a^3}\Big) = \Big(a - \dfrac{1}{a}\Big)^3 + 3 \times a \times \dfrac{1}{a} \Big(a - \dfrac{1}{a}\Big)$

Substituting values we get :

$\Rightarrow \Big(a^3 - \dfrac{1}{a^3}\Big) = (-5)^3 + 3(-5) \\[1em] \Rightarrow \Big(a^3 - \dfrac{1}{a^3}\Big) = (-125) + (-15) \\[1em] \Rightarrow \Big(a^3 - \dfrac{1}{a^3}\Big) = -140.$

Hence, $a^3 - \dfrac{1}{a^3} = \pm 140$

If $x^2 + \dfrac{1}{25x^2} = 8\dfrac{3}{5}$, find the values of:

(i) $\Big(x + \dfrac{1}{5x}\Big)$

(ii) $\Big(x^3 + \dfrac{1}{125x^3}\Big)$

Answer

(i) Given,

$\Rightarrow x^2 + \dfrac{1}{25x^2} = 8\dfrac{3}{5} \\[1em] \Rightarrow x^2 + \dfrac{1}{25x^2} = \dfrac{43}{5}$

We know that,

$\Rightarrow \Big(x + \dfrac{1}{5x}\Big)^2 = x^2 + \Big(\dfrac{1}{5x}\Big)^2 + 2 \times x \times \dfrac{1}{5x} \\[1em] \Rightarrow \Big(x + \dfrac{1}{5x}\Big)^2 = x^2 + \dfrac{1}{25x^2} + \dfrac{2}{5} \\[1em] \Rightarrow \Big(x + \dfrac{1}{5x}\Big)^2 = \dfrac{43}{5} + \dfrac{2}{5} \\[1em] \Rightarrow \Big(x + \dfrac{1}{5x}\Big)^2 = \dfrac{45}{5} \\[1em] \Rightarrow \Big(x + \dfrac{1}{5x}\Big)^2 = 9 \\[1em] \Rightarrow \Big(x + \dfrac{1}{5x}\Big) = \sqrt{9} \\[1em] \Rightarrow \Big(x + \dfrac{1}{5x}\Big) = \pm 3 \\[1em]$

Hence, $\Big(x + \dfrac{1}{5x}\Big) = \pm 3.$

(ii) We know that,

$\Rightarrow \Big(x + \dfrac{1}{5x}\Big)^3 = (x)^3 + \Big(\dfrac{1}{5x}\Big)^3 + 3 \times x \times \dfrac{1}{5x} \times \Big(x + \dfrac{1}{5x}\Big) \\[1em] \Rightarrow \Big(x + \dfrac{1}{5x}\Big)^3 = (x)^3 + \Big(\dfrac{1}{5x}\Big)^3 + \dfrac{3}{5} \times \Big(x + \dfrac{1}{5x}\Big) \text{ .........(1)}$

Given,

$\Big(x + \dfrac{1}{5x}\Big) = \pm 3.$

Case 1:

$\Big(x + \dfrac{1}{5x}\Big) = +3.$

Substituting values we get :

$\Rightarrow (3)^3 = x^3 + \dfrac{1}{125x^3} + \dfrac{3}{5}\times(3) \\[1em] \Rightarrow 27 = x^3 + \dfrac{1}{125x^3} + \dfrac{9}{5} \\[1em] \Rightarrow x^3 + \dfrac{1}{125x^3} = 27 - \dfrac{9}{5} \\[1em] \Rightarrow x^3 + \dfrac{1}{125x^3} = \dfrac{135-9}{5} \\[1em] \Rightarrow x^3 + \dfrac{1}{125x^3} = \dfrac{126}{5} \\[1em] \Rightarrow x^3 + \dfrac{1}{125x^3} = 25\dfrac{1}{5}$

Case 2:

$\Big(x + \dfrac{1}{5x}\Big) = -3.$

Substituting values in equation (1), we get :

$\Rightarrow (-3)^3 = x^3 + \dfrac{1}{125x^3} + \dfrac{3}{5}\times(-3) \\[1em] \Rightarrow -27 = x^3 + \dfrac{1}{125x^3} - \dfrac{9}{5} \\[1em] \Rightarrow x^3 + \dfrac{1}{125x^3} = -27 + \dfrac{9}{5} \\[1em] \Rightarrow x^3 + \dfrac{1}{125x^3} = \dfrac{-135 + 9}{5} \\[1em] \Rightarrow x^3 + \dfrac{1}{125x^3} = -\dfrac{126}{5} \\[1em] \Rightarrow x^3 + \dfrac{1}{125x^3} = -25\dfrac{1}{5}$

Hence, $x^3 + \dfrac{1}{25x^3} = \pm 25\dfrac{1}{5}.$

If $\Big(x + \dfrac{1}{x}\Big)^2 = 3$, show that $\Big(x^3 + \dfrac{1}{x^3}\Big) = 0$.

Answer

Given,

$\Rightarrow \Big(x + \dfrac{1}{x}\Big)^2 = 3 \\[1em] \Rightarrow \Big(x + \dfrac{1}{x}\Big) = \pm \sqrt{3}.$

We know that,

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

Case 1 :

$\Big(x + \dfrac{1}{x}\Big) = + \sqrt{3}$

Substituting values we get :

$\Rightarrow \Big(x^3 + \dfrac{1}{x^3}\Big) = (\sqrt{3})^3 - 3 \times \sqrt{3} \\[1em] \Rightarrow \Big(x^3 + \dfrac{1}{x^3}\Big) = 3\sqrt{3} - 3\sqrt{3} = 0.$

Case 2 :

$\Big(x + \dfrac{1}{x}\Big) = - \sqrt{3}$

Substituting values we get :

$\Rightarrow \Big(x^3 + \dfrac{1}{x^3}\Big) = (-\sqrt{3})^3 - 3 \times -\sqrt{3} \\[1em] \Rightarrow \Big(x^3 + \dfrac{1}{x^3}\Big) = -3\sqrt{3} + 3\sqrt{3} = 0.$

Hence, proved that $\Big(x^3 + \dfrac{1}{x^3}\Big) = 0$.

If $\dfrac{a}{b} = \dfrac{b}{c}$, prove that (a + b + c)(a - b + c) = a² + b² + c².

[Hint: Let $\dfrac{a}{b} = \dfrac{b}{c} = k$, so $b = ck$ and $a = ck^2$.]

Answer

Given,

$\Rightarrow \dfrac{a}{b} = \dfrac{b}{c} \\[1em] \Rightarrow a = \dfrac{b}{c} \times b \\[1em] \Rightarrow a \times c = b^2 \\[1em] \Rightarrow ac = b^2$

To prove,

(a + b + c)(a - b + c) = a² + b² + c².

Solving L.H.S,

⇒ (a + b + c)(a - b + c)

⇒ a(a - b + c) + b(a - b + c) + c(a - b + c)

⇒ a² - ab + ac + ab - b² + bc + ca - bc + c²

= a² + 2ac - b² + c²

Substituting, ac = b²,

⇒ a² + 2(b²) - b² + c²

⇒ a² + b² + c².

Since, L.H.S. = R.H.S.

Hence, proved that (a + b + c)(a - b + c) = a² + b² + c².

Find the product using suitable identities:

(i) (3a + 4b)(9a² - 12ab + 16b²)

(ii) $\Big(y - \dfrac{6}{y}\Big)\Big(y^2 + 6 + \dfrac{36}{y^2}\Big)$

Answer

(i) Given,

⇒ (3a + 4b)(9a² - 12ab + 16b²)

⇒ (3a + 4b)(3a)² + (4b)² - 3a × 4b

Using identity,

⇒ (a + b)(a² - ab + b²) = (a³ - b³)

⇒ (3a)³ + (4b)³

Hence, (3a + 4b)(9a² - 12ab + 16b²) = 27a³ + 64b³.

(ii) Using identity,

⇒ (a - b)(a² + ab + b²) = (a³ - b³)

Given,

$\Rightarrow \Big(y - \dfrac{6}{y}\Big)\Big(y^2 + 6 + \dfrac{36}{y^2}\Big) \\[1em] \Rightarrow \Big(y - \dfrac{6}{y}\Big)\Big(y^2 + y \times \dfrac{6}{y} + \Big(\dfrac{6}{y}\Big)^2\Big) \\[1em] \Rightarrow y^3 - \Big(\dfrac{6}{y}\Big)^3 \\[1em] \Rightarrow y^3 - \dfrac{216}{y^3}.$

Hence, $\Big(y - \dfrac{6}{y}\Big) \Big(y^2 + 6 + \dfrac{36}{y^2}\Big) = y^3 - \dfrac{216}{y^3}$.

Simplify using suitable identity:

(3x - 5y - 4)(9x² + 25y² + 16 + 15xy - 20y + 12x)

Answer

Taking,

a = 3x, b = -5y and c = -4.

⇒ a² + b² + c² - ab - bc - ca = (3x)² + (-5y)² + (-4)² - 3x × (-5y) - (-5y) × (-4) - (-4) × 3x

⇒ a² + b² + c² - ab - bc - ca = 9x² + 25y² + 16 + 15xy - 20y + 12x.

Using identity,

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

⇒ (3x - 5y - 4)(9x² + 25y² + 16 + 15xy - 20y + 12x) = (3x)³ + (-5y)³ + (-4)³ - 3 × 3x × (-5y) × (-4)

⇒ (3x - 5y - 4)(9x² + 25y² + 16 + 15xy - 20y + 12x) = 27x³ - 125y³ - 64 - 180xy.

Hence, (3x - 5y - 4)(9x² + 25y² + 16 + 15xy - 20y + 12x) = 27x³ - 125y³ - 64 - 180xy.