Simplify using following identity;

(a $\pm$ b)(a² $\mp$ ab + b²) = a³ $\pm$ b³

(i) (2x + 3y)(4x² - 6xy + 9y²)

(ii) $\Big(\dfrac{a}{3} - 3b\Big)\Big(\dfrac{a^2}{9} + ab + 9b^2\Big)$

(i) Given,

⇒ (2x + 3y)(4x² - 6xy + 9y²)

⇒ (2x + 3y)[(2x)² - 2x × 3y + (3y)²]

Comparing above equation with (a $\pm$ b)(a² $\mp$ ab $\pm$ b²) = a³ $\pm$ b³, we get :

a = 2x and b = 3y

∴ (2x + 3y)[(2x)² - 2x × 3y + (3y)²] = (2x)³ + (3y)³

= 8x³ + 27y³.

Hence, (2x + 3y)(4x² - 6xy + 9y²) = 8x³ + 27y³.

(ii) Given,

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

Using identity given in question :

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

Hence, $\Big(\dfrac{a}{3} - 3b\Big)\Big(\dfrac{a^2}{9} + ab + 9b^2\Big) = \dfrac{a^3}{27} - 27b^3$.