If x ≠ 0 and $3x + \dfrac{1}{3x} = 8$; find the value of : $27x^3 + \dfrac{1}{27x^3}$.

Given: $3x + \dfrac{1}{3x} = 8$

Using the formula: (a + b)³ = a³ + b³ + 3ab(a + b)

$\Rightarrow \Big(3x + \dfrac{1}{3x}\Big)^3 = 8^3\\[1em] \Rightarrow (3x)^3 + \Big(\dfrac{1}{3x}\Big)^3 + 3 \times 3x \times \dfrac{1}{3x} \times \Big(3x + \dfrac{1}{3x}\Big) = 512\\[1em] \Rightarrow 27x^3 + \dfrac{1}{27x^3} + 3\times 8 = 512\\[1em] \Rightarrow 27x^3 + \dfrac{1}{27x^3} + 24 = 512\\[1em] \Rightarrow 27x^3 + \dfrac{1}{27x^3} = 512 - 24\\[1em] \Rightarrow 27x^3 + \dfrac{1}{27x^3} = 488$

Hence, the value of $27x^3 + \dfrac{1}{27x^3}$ = 488.