Let alpha, alpha^2 be the roots of x^2 + x + | vedclass

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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 roots are $\alpha$

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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 roots are $\alpha$ and $\alpha^2$,so $\alpha + \alpha^2 = -1$ and $\alpha \cdot \alpha^2 = \alpha^3 = 1$.We need the equation with roots $\alpha^{31}$ and $\alpha^{62}$.Since $\alpha^3 = 1$,we have $\alpha^{31} = (\alpha^3)^{10} \cdot \alpha = 1^{10} \cdot \alpha = \alpha$.Similarly,$\alpha^{62} = (\alpha^3)^{20} \cdot \alpha^2 = 1^{20} \cdot \alpha^2 = \alpha^2$.The roots of the new equation are $\alpha$ and $\alpha^2$.Therefore,the required equation is the same as the original equation: $x^2 + x + 1 = 0$.

Let $\alpha, \alpha^2$ be the roots of $x^2 + x + 1 = 0$. Then the equation whose roots are $\alpha^{31}, \alpha^{62}$ is:

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