Factorise the following:

$x^2 - \dfrac{8}{x}$

Above terms can be written as,

$\Rightarrow x^2 - \dfrac{8}{x} = \dfrac{1}{x}\Big(x^3 - 8\Big) \\[1em] = \dfrac{1}{x}(x^3 - 2^3).$

We know that,

a³ - b³ = (a - b)(a² + ab + b²).

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

We know that,

a³ - b³ = (a - b)(a² + ab + b²).

$\dfrac{1}{x}(x^3 - 2^3) = \dfrac{1}{x}(x - 2)(x^2 + 2 \times x + 2^2) \\[1em] = \dfrac{1}{x}(x - 2)(x^2 + 2x + 4).$

Hence, $x^2 - \dfrac{8}{x} = \dfrac{1}{x}(x - 2)(x^2 + 2x + 4).$