If $x + \dfrac{1}{x} = 3\dfrac{1}{3}$, find the value of $x^3 - \dfrac{1}{x^3}$.

We know that,

$x - \dfrac{1}{x} = \sqrt{\Big(x + \dfrac{1}{x}\Big)^2 - 4}$

Substituting values we get,

$x - \dfrac{1}{x} = \sqrt{\Big(3\dfrac{1}{3}\Big)^2 - 4} \\[1em] = \sqrt{\Big(\dfrac{10}{3}\Big)^2 - 4} \\[1em] = \sqrt{\dfrac{100}{9} - 4} \\[1em] = \sqrt{\dfrac{100 - 36}{9}} \\[1em] = \sqrt{\dfrac{64}{9}} \\[1em] = \pm \dfrac{8}{3}.$

We know that,

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

Substituting values we get,

$x^3 - \dfrac{1}{x^3} = \Big(\pm \dfrac{8}{3}\Big)^3 + 3 \times \pm \Big(\dfrac{8}{3}\Big) \\[1em] = \pm \dfrac{512}{27} \pm 8 \\[1em] = \dfrac{\pm 512 \pm 216}{27} \\[1em] = \pm \dfrac{728}{27} \\[1em] = \pm 26\dfrac{26}{27}.$

Hence, the value of $x^3 - \dfrac{1}{x^3} = \pm 26\dfrac{26}{27}$.