Simplify each of the following and express it in the form (a + ib)

Given: $\left(-2-\frac{1}{3} i\right)^{3}$

We know that,

$(-a-b)^{3}=-a^{3}-3 a^{2} b-3 a b^{2}-b^{3} \ldots$ (i)

So, on replacing a by 2 and b by 1/3i in eq. (i), we get

$-(2)^{3}-3(2)^{2}\left(\frac{1}{3} i\right)-3(2)\left(\frac{1}{3} i\right)^{2}-\left(\frac{1}{3} i\right)^{3}$

$=-8-4 i-6\left(\frac{1}{9} i^{2}\right)-\left(\frac{1}{27} i^{3}\right)$

$=-8-4 i-\frac{2}{3} i^{2}-\frac{1}{27} i\left(i^{2}\right)$

$=-8-4 i-\frac{2}{3}(-1)-\frac{1}{27} i(-1)\left[\because i^{2}=-1\right]$

$=-8-4 i+\frac{2}{3}+\frac{1}{27} i$

$=\left(-8+\frac{2}{3}\right)+\left(-4 i+\frac{1}{27} i\right)$

$=\left(\frac{-24+2}{3}\right)+\left(\frac{-108 i+i}{27}\right)$

$=-\frac{22}{3}+\left(-\frac{107}{27} i\right)$

$=-\frac{22}{3}-\frac{107}{27} i$