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If $\alpha$ and $\beta$ are the roots of $x^{2}-x+1=0,$ then the equation whose roots are $\alpha^{100}$ and $\beta^{100}$ are
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Verified Answer
The correct answer is:
$x^{2}+x+1=0$
We have $x^{3}+1 \equiv(x+1)\left(x^{2}-x+1\right)$. Therefore, $\alpha$ and $\beta$ are the complex cube roots of -1 so that we may take $\alpha=-\omega$ and $\beta=-\omega^{2},$ where $\omega \neq 1$ is a cube root of unity.
Thus $\alpha^{100}=(-\omega)^{100}=\omega$ and
$\beta^{100}=\left(-\omega^{2}\right)^{100}=\omega^{2},$ so that the required
equation is $\mathrm{x}^{2}+\mathrm{x}+1=0$
Thus $\alpha^{100}=(-\omega)^{100}=\omega$ and
$\beta^{100}=\left(-\omega^{2}\right)^{100}=\omega^{2},$ so that the required
equation is $\mathrm{x}^{2}+\mathrm{x}+1=0$
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