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

$\begin{array}{rcl}-\mathrm{i}& =& \cos \frac{\mathrm{\pi }}{2}-\mathrm{i} \sin \frac{\mathrm{\pi }}{2}\end{array}$

$\begin{array}{rcl}{(-\mathrm{i})}^{1/3}& =& {\left(\cos \frac{\mathrm{\pi }}{2}-\mathrm{i} \sin \frac{\mathrm{\pi }}{2}\right)}^{1/3}\end{array}$

$\begin{array}{rcl} & =& \cos \frac{{\displaystyle \frac{\mathrm{\pi }}{2}+2\mathrm{k\pi }}}{3}-\mathrm{i} \sin \frac{2\mathrm{k\pi }+{\displaystyle \frac{\mathrm{\pi }}{2}}}{3}; \mathrm{k}=0, 1, 2\end{array}$

$\begin{array}{rcl}\mathrm{\alpha }& =& \cos \frac{\mathrm{\pi }}{6}-\mathrm{i} \sin \frac{\mathrm{\pi }}{6}, \mathrm{\beta }=\cos \frac{5\mathrm{\pi }}{6}-\mathrm{i} \sin \frac{5\mathrm{\pi }}{6},\end{array}$

$\begin{array}{rcl}\mathrm{\gamma }& =& \cos \frac{9\mathrm{\pi }}{6}-\mathrm{i} \sin \frac{9\mathrm{\pi }}{6}\end{array}$

$\begin{array}{rcl} & & {\mathrm{\alpha }}^2+{\mathrm{\beta }}^2+{\mathrm{\gamma }}^2\end{array}$

$\begin{array}{rcl} & =& {\left(\cos \frac{\mathrm{\pi }}{6}-\mathrm{i} \sin \frac{\mathrm{\pi }}{6}\right)}^2+{\left(\cos \frac{5\mathrm{\pi }}{6}-\mathrm{i} \sin \frac{5\mathrm{\pi }}{6}\right)}^2+{\left(\cos \frac{3\mathrm{\pi }}{2}-\mathrm{i} \sin \frac{3\mathrm{\pi }}{2}\right)}^2\end{array}$

$\begin{array}{rcl} & =& \cos \frac{\mathrm{\pi }}{3}-\mathrm{i} \sin \frac{\mathrm{\pi }}{3}+\cos \frac{5\mathrm{\pi }}{3}-\mathrm{i} \sin \frac{5\mathrm{\pi }}{3}+{\left(\mathrm{i}\right)}^2 \end{array}$

$\begin{array}{rcl} & =& \end{array}\begin{array}{rcl} & & \frac{1}{2}\end{array}\begin{array}{rcl} & & -\end{array}\begin{array}{rcl} & & \frac{\mathrm{i}\sqrt{3}}{2}\end{array}\begin{array}{rcl} & & +\end{array}\begin{array}{rcl} & & \frac{1}{2}\end{array}\begin{array}{rcl} & & +\end{array}\begin{array}{rcl} & & \frac{\mathrm{i}\sqrt{3}}{2}\end{array}\begin{array}{rcl} & & -\end{array}\begin{array}{rcl} & & 1\end{array}=0$