Simplify the following:

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

The above equation can be written as,

$\Rightarrow \Big(x^{\dfrac{1}{3}} - x^{-\dfrac{1}{3}}\Big)\Big[\Big(x^{\dfrac{1}{3}}\Big)^2 + x^{\dfrac{1}{3}}.x^{-\dfrac{1}{3}} + \Big(x^{-\dfrac{1}{3}}\Big)^2\Big)$

We know that,

a³ - b³ = (a - b)(a² + ab + b²)

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

Hence, $\Big(x^{\dfrac{1}{3}} - x^{-\dfrac{1}{3}}\Big)\Big(x^{\dfrac{2}{3}} + 1 + x^{-\dfrac{2}{3}}\Big) = x - \dfrac{1}{x}$.