Explanation

(a)
The given equation $\mathcal{z}^{3} + 2\mathcal{z}^{2} + 2\mathcal{z} + 1 = 0$ can be rewritten as $\left( \mathcal{z} + 1 \right)\left( \mathcal{z}^{2}\mathcal{+ z} + 1 \right) = 0.$ Its roots $- 1,\omega\ and\ \omega^{2}$
Let $f\left( \mathcal{z} \right) = \mathcal{z}^{1985} + \mathcal{z}^{100} + 1$
Put $\mathcal{z} = - 1,\ \omega$ and $\omega^{2}$ respectively, we have
$\ f( - 1) = ( - 1)^{1985} + ( - 1)^{100} + 1 \neq 0$
Therefore, $- 1$ is not a root of the equation $f(\mathcal{z}) = 0$
Again, $f(\omega) = \omega^{1985} + \omega^{100} + 1$
$= \left( \omega^{3} \right)^{661}\omega^{2} + \left( \omega^{3} \right)^{33}\omega + 1$
$= \omega^{2} + \omega + 1 = 0$
Therefore, $\omega$ is a root of the equation $f(\mathcal{z)}$=0
Similarly, $f\left( \omega^{2} \right) = 0$
Hence, $\omega$ and $\omega^{2}$ are the common roots