$\alpha $ and $\alpha^2 $ are the roots of the equation $x^2+x+1=0$ then the equation whose roots are $\alpha^{31} $ and $\alpha^{62} $ is: $$a). x^2+x+1=0$$ $$b). x^2-x+1=0$$ $$c). x^2-x-1=0$$ $$d). x^2-2x-1=0$$
My Attempt:
Since, $\alpha $ and $\alpha^2 $ are the roots of $x^2+x+1=0$ so, $$\alpha + \alpha^2 = -1$$ $$\alpha.\alpha^2=1$$ ...
You just found out that $\alpha^3=1$. Then, we get, $$\alpha^{31}=(\alpha^3)^{10}\alpha =\alpha \text{ and } \alpha^{62}=(\alpha^3)^{20}\alpha^2=\alpha^2$$
Thus, $\alpha^{31}$ and $\alpha^{62}$ satisfy the equation: $$x^2+x+1=0$$