If $x - \dfrac{1}{x} = 3 + 2\sqrt{2}$, find the value of $\dfrac{1}{4}\Big(x^3 - \dfrac{1}{x^3}\Big)$.

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,

$x^3 - \dfrac{1}{x^3} = (3 + 2\sqrt{2})^3 + 3(3 + 2\sqrt{2}) \\[1em] = (3)^3 + (2\sqrt{2})^3 + 3(3)(2\sqrt{2})(3 + 2\sqrt{2}) + 9 + 6\sqrt{2} \\[1em] = 27 + 16\sqrt{2} + 18\sqrt{2}(3 + 2\sqrt{2}) + 9 + 6\sqrt{2} \\[1em] = 27 + 9 + 16\sqrt{2} + 54\sqrt{2} + 72 + 6\sqrt{2} \\[1em] = 108 + 76\sqrt{2}.$

So, $\dfrac{1}{4}\Big(x^3 - \dfrac{1}{x^3}\Big) = \dfrac{1}{4} \times (108 + 76\sqrt{2}) = 27 + 19\sqrt{2}$.

Hence, $\dfrac{1}{4}\Big(x^3 - \dfrac{1}{x^3}\Big) = 27 + 19\sqrt{2}$.