Expand the following:
$\left(4-\frac{1}{3 x}\right)^{3}$

(D) We use the algebraic identity $(a-b)^{3} = a^{3} - b^{3} - 3ab(a-b)$.
Here,$a = 4$ and $b = \frac{1}{3x}$.
Substituting these values into the identity:
$\left(4-\frac{1}{3 x}\right)^{3} = (4)^{3} - \left(\frac{1}{3 x}\right)^{3} - 3(4)\left(\frac{1}{3 x}\right)\left(4-\frac{1}{3 x}\right)$
$= 64 - \frac{1}{27 x^{3}} - \frac{12}{3 x}\left(4-\frac{1}{3 x}\right)$
$= 64 - \frac{1}{27 x^{3}} - \frac{4}{x}\left(4-\frac{1}{3 x}\right)$
$= 64 - \frac{1}{27 x^{3}} - \frac{16}{x} + \frac{4}{3 x^{2}}$