Factorise :

$\dfrac{1}{3}x^2 - \dfrac{8}{x}$

$\dfrac{1}{3}x^2 - \dfrac{8}{x}\\[1em] =\dfrac{x^3 - 24}{3x}\\[1em] = \dfrac{x^3 - (2\sqrt[3]{3})^3}{3x}\\[1em]$

Using the formula: (a³ - b³) = (a - b)(a² + ab + b²)

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

Hence, $\dfrac{1}{3}x^2 - \dfrac{8}{x} = \dfrac{(x - 2\sqrt[3]{3})(x^2 + x2\sqrt[3]{3} + 4(\sqrt[3]{3})^2)}{3x}$.