If the complex cube roots of$(-\mathrm{i})$ are $\alpha , \beta , \gamma ,$ then ${\alpha }^2+{\beta }^2+{\gamma }^2=$

${\left[-i\right]}^{\frac{1}{3}}=\alpha , \beta , \gamma$

${\alpha }^2+{\beta }^2+{\gamma }^2=?$

$\left|-i\right|=1$ and argument is =$\left[-\frac{\mathrm{\pi }}{2}\right]$

$\Rightarrow$So we can change it in polar form

${\left[\cos \left(2m\mathrm{\pi }-\frac{\mathrm{\pi }}{2}\right)+i\sin \left(2m\mathrm{\pi }-\frac{\mathrm{\pi }}{2}\right)\right]}^{\frac{1}{3}}$

$\Rightarrow$$\cos \left(\frac{2m\mathrm{\pi }}{3}-\frac{\mathrm{\pi }}{6}\right)+i\sin \left[\frac{2m\mathrm{\pi }}{3}-\frac{\mathrm{\pi }}{6}\right]$

putting $m=0, 1, 2$

$\Rightarrow \cos \left(-\frac{\mathrm{\pi }}{6}\right)+i\sin \left(-\frac{\mathrm{\pi }}{6}\right)$

$\Rightarrow$$\frac{\sqrt{3}}{2}-i\frac{1}{2}=\alpha$

Now putting $m=1$

$\cos \left(\frac{2\mathrm{\pi }}{3}-\frac{\mathrm{\pi }}{6}\right)+i\sin \left(\frac{2\mathrm{\pi }}{3}-\frac{\mathrm{\pi }}{6}\right)$

$\Rightarrow$$\cos \frac{\mathrm{\pi }}{2}+i\sin \frac{\mathrm{\pi }}{2}$$\Rightarrow$$i=\beta$

Now putting $m=2$

$\cos \left(\frac{4\mathrm{\pi }}{3}-\frac{7}{6}\right)+i\sin \left(\frac{4\mathrm{\pi }}{3}-\frac{\mathrm{\pi }}{6}\right)$

$\Rightarrow \cos \frac{7\mathrm{\pi }}{6}+i\sin \frac{7\mathrm{\pi }}{6}$

$\Rightarrow \cos \left(\mathrm{\pi }+\frac{\mathrm{\pi }}{6}\right)+i\sin \left(\mathrm{\pi }+\frac{\mathrm{\pi }}{6}\right)$

$\Rightarrow$$-\cos \frac{\mathrm{\pi }}{6}-i\sin \frac{\mathrm{\pi }}{6}$

$\Rightarrow$$-\frac{\sqrt{3}}{2}-i\frac{1}{2}=\gamma$

$\Rightarrow$So, ${\alpha }^2+{\beta }^2+{\gamma }^2=$$\left[\frac{3}{4}-\frac{1}{4}-\cancel{\frac{\sqrt{3}}{2}i}\right]-1+\left[\frac{3}{4}-\frac{1}{4}+\cancel{\frac{\sqrt{3}}{2}i}\right]=0$