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

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