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 
$$x_1^8+x_2^8+\cdots+x_8^8$$

Use Newton's identity, or a particular case of it, suited to the particular polynomial at hand. You have for all $i$
$$
x_i^8-13x_i^2+7x_i-6=0.
$$
Sum over $i$ to get
$$
S - 13 Q -48 = 0,
$$
where $S = x_1^8+x_2^8+\cdots+x_8^8$ is your sum, $Q = x_1^2+x_2^2+\cdots+x_8^2$, and we have used the fact that in a polynomial $$x^8 - a_1 x^7 + a_2 x^6 - \dots,\tag{pol}$$ the coefficient $a_1$ is the sum of the roots.
Also, in (pol) $a_2$ is the sum of the $x_i x_j$ for $i < j$. Now $Q = a_1^2 - 2 a_2 = 0$ here.
So $S = 48$.
