What is the value of $\alpha^{8}+\beta^{8}+\gamma^{8}$ if $\alpha$, $\beta$ and $\gamma$ are roots of the equation $x^3+x-1$? Is there a shorter way of finding the answer apart from finding the individual values of the roots.
The value of $\alpha^{8}+\beta^{8}+\gamma^{8}$ if $\alpha$, $\beta$ and $\gamma$ are roots of the equation $x^3+x-1$
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Best Answer
$a,b,c$ be the roots of $x^3+x-1=0$, then $a^3=1-a \implies a^8=\frac{(1-a)^3}{a}=\frac{1-a^3-3a+3a^2}{a}=\frac{1-(1-a)-3a+3a^2}{a}$ $$\implies a^8=3a-2,\implies a^8+b^8+c^8=3(a+b+c)-6=-6$$ as the sum of the roots is zero.