Explanation

(b)
Given, $z^{2} + z + 1 = 0$
$\Rightarrow \ \ \ \ \ z = \omega,\ \omega^{2}$
Take $z = \omega$
$\therefore\ \ \ \ \left( z + \frac{1}{z} \right)^{2} + \left( z^{2} + \frac{1}{z^{2}} \right)^{2} + \left( z^{3} + \frac{1}{z^{3}} \right)^{2} + \left( z^{4} + \frac{1}{z^{4}} \right)^{2} + \left( z^{5} + \frac{1}{z^{5}} \right)^{2} + \left( z^{6} + \frac{1}{z^{6}} \right)^{2}$
$= \left( \omega + \frac{1}{\omega} \right)^{2} + \left( \omega^{2} + \frac{1}{\omega^{2}} \right)^{2} + \left( \omega^{3} + \frac{1}{\omega^{3}} \right)^{2} + \left( \omega^{4} + \frac{1}{\omega^{4}} \right)^{2} + \left( \omega^{5} + \frac{1}{\omega^{5}} \right)^{5} + \left( \omega^{6} + \frac{1}{\omega^{6}} \right)^{2}$
$= \left( \omega + \omega^{2} \right)^{2} + \left( \omega^{2} + \omega \right)^{2} + (1 + 1)^{2} + \left( \omega + \omega^{2} \right)^{2} + \left( \omega^{2} + \omega \right)^{2} + (1 + 1)^{2}$
$= 1 + 1 + 4 + 1 + 1 + 4 = 12$
Similarly, for $z = \omega^{2}$, we get the same result