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 $\alpha$ and $\alpha^2$ are roots,we know $\alpha + \alpha^2 = -1$ and $\alpha^3 = 1$.We need to find t

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$x^2 + x + 1 = 0$

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(C) Given the equation $x^2 + x + 1 = 0$.Since $\alpha$ and $\alpha^2$ are roots,we know $\alpha + \alpha^2 = -1$ and $\alpha^3 = 1$.We need to find the equation with roots $\alpha^{31}$ and $\alpha^{62}$.Sum of roots: $\alpha^{31} + \alpha^{62} = \alpha^{31}(1 + \alpha^{31}) = (\alpha^3)^{10} \cdot \alpha (1 + (\alpha^3)^{10} \cdot \alpha) = \alpha(1 + \alpha) = \alpha + \alpha^2 = -1$.Product of 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 $x^2 + x + 1 = 0$.Alternatively,since $\alpha$ is a complex cube root of unity $\omega$,then $\alpha^{31} = \omega^{31} = \omega$ and $\alpha^{62} = \omega^{62} = \omega^2$. The equation with roots $\omega$ and $\omega^2$ is $x^2 + x + 1 = 0$.

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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