Evaluate:

$\Big[\Big(\dfrac{1}{4}\Big)^{-3} - \Big(\dfrac{1}{3}\Big)^{-3}\Big] ÷ \Big(\dfrac{1}{6}\Big)^{-3}$

As we know, for any non-zero rational number a

$a^{-n} = \dfrac{1}{a^n}$ and $a^{n} = \dfrac{1}{a^{-n}}$.

$\Big[\Big(\dfrac{1}{4}\Big)^{-3} - \Big(\dfrac{1}{3}\Big)^{-3}\Big] ÷ \Big(\dfrac{1}{6}\Big)^{-3}\\[1em] = \Big[\Big(\dfrac{4}{1}\Big)^3 - \Big(\dfrac{3}{1}\Big)^3\Big] ÷ \Big(\dfrac{6}{1}\Big)^3\\[1em] = [(4 \times 4 \times 4) - (3 \times 3 \times 3)] ÷ (6 \times 6 \times 6)\\[1em] = [64 - 27] ÷ (216)\\[1em] = [37] ÷ (216)\\[1em] = [37] \times \Big(\dfrac{1}{216}\Big)\\[1em] = \Big(\dfrac{37 \times 1}{216}\Big)\\[1em] = \Big(\dfrac{37}{216}\Big)$

Hence, $\Big[\Big(\dfrac{1}{4}\Big)^{-3} - \Big(\dfrac{1}{3}\Big)^{-3}\Big] ÷ \Big(\dfrac{1}{6}\Big)^{-3} = \dfrac{37}{216}$.