The value of ${(-i)}^{1/3}$ is

Since  $\frac{-\sqrt{3}-i}{2}=-\left(\cos \frac{{\displaystyle \pi }}{{\displaystyle 6}}+i\sin \frac{{\displaystyle \pi }}{{\displaystyle 6}}\right)$

$\Rightarrow {\left(\frac{{\displaystyle -\sqrt{3}-i}}{{\displaystyle 2}}\right)}^3=-{\left(\cos \frac{{\displaystyle \pi }}{{\displaystyle 6}}+i\sin \frac{{\displaystyle \pi }}{{\displaystyle 6}}\right)}^3=-i$

and $\frac{\sqrt{3}-i}{2}=\cos \frac{\pi }{6}-i\sin \frac{\pi }{6}$

and ${\left(\frac{{\displaystyle \sqrt{3}-i}}{{\displaystyle 2}}\right)}^3=\cos \frac{\pi }{2}-i\sin \frac{\pi }{2}=-i$

Hence the result.