$\Big(\dfrac{x^3}{8} - \dfrac{y^3}{27}\Big)$ in the form of factors is :

1. $\Big(\dfrac{x}{2} - \dfrac{y}{3}\Big)\Big(\dfrac{x^2}{4} - \dfrac{xy}{6} + \dfrac{y^2}{9}\Big)$

2. $\Big(\dfrac{x}{2} + \dfrac{y}{3}\Big)\Big(\dfrac{x^2}{4} - \dfrac{xy}{6} + \dfrac{y^2}{9}\Big)$

3. $\Big(\dfrac{x}{2} - \dfrac{y}{3}\Big)\Big(\dfrac{x^2}{4} + \dfrac{xy}{6} + \dfrac{y^2}{9}\Big)$

4. $\Big(\dfrac{x}{2} + \dfrac{y}{3}\Big)\Big(\dfrac{x^2}{4} + \dfrac{xy}{6} + \dfrac{y^2}{9}\Big)$

Given,

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

Hence, Option 3 is the correct option.