Explanation

(c)
Since, $\alpha,\ \alpha^{2}$ be the roots of $x^{2} + x + 1 = 0$.
$\therefore\ \alpha + \alpha^{2} = - 1$ ...(i)
and $\alpha^{3} = 1$ ...(ii)
Now, $\alpha^{31} + \alpha^{62} = \alpha^{31}\left( 1 + \alpha^{31} \right)$
$\Rightarrow \ \alpha^{31} + \alpha^{62} = \alpha^{30}(1 + \alpha^{30} \bullet \alpha)$
$\Rightarrow \ \alpha^{31} + \alpha^{62} = \left( \alpha^{3} \right)^{10} \bullet \alpha\{ 1 + \left( \alpha^{3} \right)^{10} \bullet \alpha\}$
$\Rightarrow \ \alpha^{31} + \alpha^{62} = \alpha(1 + \alpha)$ [from Eq. (ii)]
$\Rightarrow \ \alpha^{31} + \alpha^{62} = - 1$ [from Eq. (i)]
Again $\alpha^{31} \bullet \alpha^{62} = \alpha^{93}$
$\alpha^{31} \bullet \alpha^{62} = \left\lbrack \alpha^{3} \right\rbrack^{31} = 1$
$\therefore$ Required equation is
$x^{2} - \left( \alpha^{31} + \alpha^{62} \right)x + \alpha^{31} \bullet \alpha^{62} = 0$
$\Rightarrow x^{2} + x + 1 = 0$