$\Big(\dfrac{1}{3}\Big)^{-3} - \Big(\dfrac{1}{2}\Big)^{-3}$ is equal to :

1. 1

2. $\dfrac{1}{27} - \dfrac{1}{8}$

3. 19

4. -19

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

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

Hence,

$\Big(\dfrac{1}{3}\Big)^{-3} - \Big(\dfrac{1}{2}\Big)^{-3}\\[1em] = \Big(\dfrac{3}{1}\Big)^3 - \Big(\dfrac{2}{1}\Big)^3\\[1em] = \Big(\dfrac{3 \times 3 \times 3}{1 \times 1 \times 1}\Big) - \Big(\dfrac{2 \times 2 \times 2}{1 \times 1 \times 1}\Big) \\[1em] = \dfrac{27}{1} - \dfrac{8}{1} \\[1em] = 19$

Hence, option 3 is the correct option.