If ${\mathrm{\alpha }}_1,{\mathrm{\alpha }}_2,{\mathrm{\alpha }}_3-----{\mathrm{\alpha }}_{30}$ are 30^th roots of unity then $\sum \sum _{1\le \mathrm{i}<\mathrm{j}\le 30} {\left({\mathrm{\alpha }}_{\mathrm{i}}{\mathrm{\alpha }}_{\mathrm{j}}\right)}^3=$

$\begin{array}{l}\sum _{1\le \mathrm{i}<\mathrm{j}\le 30} \sum _{\mathrm{i}}^3 {\mathrm{\alpha }}_{\mathrm{j}}^3=\frac{\left({\mathrm{\alpha }}_1^3+{\mathrm{\alpha }}_2^3+---+{{\mathrm{\alpha }}_{30}}^3\right)\left({\mathrm{\alpha }}_1^6+{{\mathrm{\alpha }}_2}^6+---+{{\mathrm{\alpha }}_{30}}^6\right)}{2}\\ {\mathrm{\alpha }}_1^3+{{\mathrm{\alpha }}_2}^3+---+{{\mathrm{\alpha }}_{30}}^3=1+(\mathrm{\alpha }{)}^3+{\left({\mathrm{\alpha }}^2\right)}^3+-----+{\left({\mathrm{\alpha }}^{29}\right)}^3\left(\mathrm{\alpha }=\mathrm{cis}\frac{2\mathrm{\pi }}{30}\right)\\ =\frac{1-{\left({\mathrm{\alpha }}^3\right)}^{30}}{1-{\mathrm{\alpha }}^3}=0\\ \text{ Similarly }{{\mathrm{\alpha }}_1}^6+{{\mathrm{\alpha }}_2}^6+----+{{\mathrm{\alpha }}_{30}}^6=0\end{array}$