Factorise:

$216x^3 + \dfrac{1}{27}$

Given,

⇒ $216x^3 + \dfrac{1}{27}$

By using the identity,

a³ + b³ = (a + b)(a² - ab + b²)

$\Rightarrow (6x)^3 + \Big(\dfrac{1}{3}\Big)^3 \\[1em] \Rightarrow \Big(6x + \dfrac{1}{3}\Big)\Big[(6x)^2 - 6x \times \Big(\dfrac{1}{3}\Big) + \Big(\dfrac{1}{3}\Big)^2\Big] \\[1em] \Rightarrow \Big(6x + \dfrac{1}{3}\Big)\Big(36x^2 - 2x + \dfrac{1}{9}\Big).$

Hence, $216x^3 + \dfrac{1}{27} = \Big(6x + \dfrac{1}{3}\Big)\Big(36x^2 - 2x + \dfrac{1}{9}\Big)$.