The complex number

$\frac{(1-i)^{3}}{1-i^{3}}=\frac{(1-i)^{3}}{1+i} \quad\left(\because i^{2}=-1\right)$

$=\frac{(1-i)^{3}}{1+i} \times \frac{1-i}{1-i}$

$=\frac{(1-i)^{4}}{1-i^{2}}$

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

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

$=\frac{1}{2}\{(1-1-2 i)(1-1-2 i)\}$

$=\frac{1}{2}\{+2 i \times 2 i\}$

$=\frac{1}{2} \times 4 i^{2}$

$=-2=-2(\cos 0+i \sin 0)$

$\because \cos 0=1$ and $\sin 0=0$

Hence $\frac{(1-i)^{3}}{1-i^{3}}=2(\cos \pi+i \sin \pi)$