<s>[INST] Finding $x^8+y^8+z^8$, given $x+y+z=0$, $\;xy +xz+yz=-1$, $\;xyz=-1$ 
The system says
$$x+y+z=0$$
$$xy +xz+yz=-1$$
$$xyz=-1$$
Find
$$x^8+y^8+z^8$$

With the first equation I squared and I found that $$x^2+y^2+z^2 =2$$
trying with
$$(x + y + z)^3 = 
 x^3 + y^3 + z^3 +  3 x^2 y + 3 x y^2 + 3 x^2 z++ 3 y^2 z + 3 x z^2 + 
  3 y z^2 + 6 x y z$$
taking advantage of the fact that there is an $xyz=-1$ in the equation, but I'm not getting anywhere, someone less myopic than me.how to solve it?
Thanks
Edit : Will there be any university way to solve this problem , they posed it to a high school friend and told him it was just manipulations of remarkable products.
His answers I understand to some extent but I don't think my friend understands all of it.
 [/INST] $x, y$ and $z$ are roots of the polynomial $x^3-x+1=0$
In the usual notation ,we have $$\Sigma x^2=(\Sigma x)^2-2\Sigma xy=0-2(-1)=2$$
Then since the relation $x^3=x-1$ holds for all three roots, summing these gives:
$$\Sigma x^3=\Sigma x - \Sigma 1=0-3=-3$$
Likewise,
$$\Sigma x^4=\Sigma x^2-\Sigma x=2$$
$$\Sigma x^5=\Sigma x^3-\Sigma x^2=-3-2=-5$$
$$\Sigma x^6=\Sigma x^4-\Sigma x^3=2--3=5$$
And finally,
$$\Sigma x^8=\Sigma x^6-\Sigma x^5=5--5=10$$
</s>