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

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

Solution 1:

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 $S = 48$.

Solution 2:

$x^8-13x^2+7x-6=\prod_{k=1}^8(x-x_k)$, 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,

$$\begin{align*} \sum_{k=1}^8x_k^8&=\sum_{k=1}^8\left(13x_k^2-7x_k+6\right)\\ &=13\sum_{k=1}^8x_k^2-7\sum_{k=1}^8x_k+48\\ &=48\;. \end{align*}$$

Solution 3:

If $r_i, 1\le i\le 8$ are the roots of the given equation

applying Vieta's Formulas, $S_1=\sum r_i=0,\sum_{i\ne j}r_ir_j=0$ and so on.

Now apply Newton's formula of sum of powers of roots,

if $S_n=\sum{r^n} , S_{n+8}-13S_{n+2}+7S_{n+1}-6S_{n}=0$ if $n\ge 0$ else consider the coefficient of $S_n$ to be $0$

For example, $S_0=8$