What is the value of $\alpha^{8}+\beta^{8}+\gamma^{8}$ if $\alpha$, $\beta$ and $\gamma$ are roots of the equation $x^3+x-1$? 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.
 A: $a,b,c$ 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.
A: For all roots,
$$x^3=1-x,$$
and
$$x^8=\frac{x^9}x=\frac{1-3x+3x^2-x^3}x=3x-2.$$
Then using Vieta,
$$S_8=3S_1-3\cdot2=0-6.$$
A: If $\alpha$ is a root then $\alpha^3=1-\alpha$, so $\alpha^8=(\alpha^3)^2\alpha^2=(1-\alpha)^2\alpha^2=\alpha^4-2\alpha^3+\alpha^2$. Now you can reduce the degree of $\alpha^4$ and $\alpha^3$ in the same way to obtain an expression that has no powers larger than 1.
Apply the same reasoning to $\beta$ and $\gamma$ and use Vieta's formula for the sum of roots.
A: You have\begin{align}\alpha^8&=\alpha^2\left(\alpha^3\right)^2\\&=\alpha^2(-\alpha+1)^2\\&=\alpha^4-2\alpha^3+\alpha^2\\&=\alpha(-\alpha+1)-2(-\alpha+1)+\alpha^2\\&=3\alpha-2.\end{align}Can you take it from here?
A: The systematic way that does not need any insight is to use polynomial division:
$$
x^8=(x^5 - x^3 + x^2 + x - 2)(x^3+x-1)+( 3 x - 2)
$$
The quotient is not important; the remainder is. It tells us that $\alpha^8=3\alpha-2$ and the same for $\beta$ and $\gamma$. Then we can use  Vieta's formula for the sum of roots.
Another approach is to use Newton's identities.
