If $x - \dfrac{2}{x} = 3$, find the value of $x^3 - \dfrac{8}{x^3}$.

We know that,

⇒ (a - b)³ = a³ - b³ - 3ab(a - b)

⇒ a³ - b³ = (a - b)³ + 3ab(a - b).

$x^3 - \dfrac{8}{x^3} = (x)^3 - \Big(\dfrac{2}{x}\Big)^3 \\[1em] = \Big(x - \dfrac{2}{x}\Big)^3 + 3 \times x \times \dfrac{2}{x}\Big(x - \dfrac{2}{x}\Big) \\[1em] = \Big(x - \dfrac{2}{x}\Big)^3 + 6\Big(x - \dfrac{2}{x}\Big).$

Substituting values we get,

$x^3 - \dfrac{8}{x^3} = 3^3 + 6 \times 3 = 27 + 18 = 45.$

Hence, the value of $x^3 - \dfrac{8}{x^3} = 45.$