$1,{\mathrm{\alpha }}_1,{\mathrm{\alpha }}_2,{\mathrm{\alpha }}_3\dots \dots \dots {\mathrm{\alpha }}_{\mathrm{n}-1}$ are the $n^{\mathrm{th}}$ roots of unity and n is an even natural number, then $\left(1+{\mathrm{\alpha }}_1\right)\left(1+{\mathrm{\alpha }}_2\right)\left(1+{\mathrm{\alpha }}_3\right)\dots \dots \left(1+{\mathrm{\alpha }}_{\mathrm{n}-1}\right)=$

$\mathrm{n}=$even natural number.

$1,{\mathrm{\alpha }}_1,{\mathrm{\alpha }}_2,{\mathrm{\alpha }}_3\dots \dots \dots {\mathrm{\alpha }}_{\mathrm{n}-1}$ are ${\mathrm{n}}^{\mathrm{th}}$ roots of unity.

${\mathrm{x}}^{\mathrm{n}}-1=(\mathrm{x}-1) (\mathrm{x}-{\mathrm{\alpha }}_1) (\mathrm{x}-{\mathrm{\alpha }}_2)....(\mathrm{x}-{\mathrm{\alpha }}_{\mathrm{n}-1})$

Let  $\mathrm{x}=-1$

$\begin{array}{rcl}{(-1)}^{\mathrm{n}}-1& =& (-2) (-1-{\mathrm{\alpha }}_1) (-1-{\mathrm{\alpha }}_2)...(-1-{\mathrm{\alpha }}_{\mathrm{n}-1})\end{array}$

$\begin{array}{rcl}1-1& =& 2(1+{\mathrm{\alpha }}_1) (1+{\mathrm{\alpha }}_2)...(1+{\mathrm{\alpha }}_{\mathrm{n}-1})\end{array}$

$\begin{array}{rcl} & & \\ \therefore (1+{\mathrm{\alpha }}_1) (1+{\mathrm{\alpha }}_2)....(1+{\mathrm{\alpha }}_{\mathrm{n}-1})& =& 0\end{array}$