Finding :$x_1^8+x_2^8+\cdots+x_8^8$ If $x_1,x_2,\ldots,x_8$ are roots for the equation :
$$x^8-13x^2+7x-6=0$$ then how to find 
$$x_1^8+x_2^8+\cdots+x_8^8$$

$x^8-13x^2+7x-6=\prod_{k=1}^8(x-x_k)$, in which the coefficient of $x^7$ is $-\sum_{k=1}^8x_k$, so $\sum_{k=1}^8x_k=0$. The coefficient of $x^6$ is $\sum_{1\le i<k\le 8}x_ix_k$, so
$$0=\left(\sum_{k=1}^8x_k\right)^2=\sum_{k=1}^8x_k^2+2\sum_{1\le i<k\le 8}x_ix_k=\sum_{k=1}^8x_k^2+0\;,$$
and $\sum_{k=1}^8x_k^2=0$. Thus,
$$\begin{align*}
\sum_{k=1}^8x_k^8&=\sum_{k=1}^8\left(13x_k^2-7x_k+6\right)\\
&=13\sum_{k=1}^8x_k^2-7\sum_{k=1}^8x_k+48\\
&=48\;.
\end{align*}$$
