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(i) (3a + 5b)3

(ii) (2p - 3q)3

(iii) (2x+13x)3\Big(2x + \dfrac{1}{3x}\Big)^3

(iv) (3ab - 2c)3

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

(vi) (12x−23y)3\Big(\dfrac{1}{2}x - \dfrac{2}{3}y\Big)^3

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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,

(2x+13x)3\Big(2x + \dfrac{1}{3x}\Big)^3

Using identity :

(a + b)3 = a3 + b3 + 3a2b + 3ab2

⇒(2x+13x)3=[(2x)3+(13x)3+3×(2x)2×(13x)+3×2x×(13x)2]⇒(2x+13x)3=8x3+127x3+3×4x2×(13x)+3×2x×(19x2)⇒(2x+13x)3=8x3+4x+23x+127x3\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, (2x+13x)3=8x3+4x+23x+127x3\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,

(3a−1a)3\Big(3a - \dfrac{1}{a}\Big)^3

Using identity :

(a - b)3 = a3 - b3 - 3a2b + 3ab2

⇒(3a−1a)3=[(3a)3−(1a)3−3×(3a)2×(1a)+3×3a×(1a)2]⇒(3a−1a)3=27a3−1a3−3×9a2×(1a)+3×3a×(1a2)⇒(3a−1a)3=27a3−27a+9a−1a3\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, (3a−1a)3=27a3−27a+9a−1a3\Big(3a - \dfrac{1}{a}\Big)^3 = 27a^3 - 27a + \dfrac{9}{a} - \dfrac{1}{a^3}.

(vi) Given,

(12x−23y)3\Big(\dfrac{1}{2}x - \dfrac{2}{3}y\Big)^3

Using identity :

(a - b)3 = a3 - b3 - 3a2b + 3ab2

⇒(12x−23y)3=[(12x)3−(23y)3−3×(12x)2×(23y)+3×(12x)×(23y)2]⇒(12x−23y)3=(18x3)−(827y3)−3×(14x2)×(23y)+3×(12x)×(49y2)⇒(12x−23y)3=18x3−827y3−612x2y+1218xy2⇒(12x−23y)3=18x3−827y3−12x2y+23xy2\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, (12x−23y)3=18x3−827y3−12x2y+23xy2\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.

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