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(C) Given the equation $x^2 + x + 1 = 0$. Since $\alpha$ is a root,$\alpha^2 + \alpha + 1 = 0$,which implies $\alpha^3 = 1$. Also,the sum of roots $\a

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$x^2 + x + 1 = 0$

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(C) Given the equation $x^2 + x + 1 = 0$. Since $\alpha$ is a root,$\alpha^2 + \alpha + 1 = 0$,which implies $\alpha^3 = 1$. Also,the sum of roots $\alpha + \alpha^2 = -1$. We need to find the equation with roots $\alpha^{31}$ and $\alpha^{62}$. Sum of new roots: $\alpha^{31} + \alpha^{62} = \alpha^{30} \cdot \alpha + (\alpha^3)^{20} \cdot \alpha^2 = (1)^{10} \cdot \alpha + (1)^{20} \cdot \alpha^2 = \alpha + \alpha^2 = -1$. Product of new roots: $\alpha^{31} \cdot \alpha^{62} = \alpha^{93} = (\alpha^3)^{31} = 1^{31} = 1$. The required quadratic equation is $x^2 - (	ext{sum of roots})x + (	ext{product of roots}) = 0$. Substituting the values: $x^2 - (-1)x + 1 = 0$,which simplifies to $x^2 + x + 1 = 0$.

If $\alpha$ 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 .....

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