<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] You got already that $$x^2+y^2+z^2=2.$$
Also, we have:
$$\sum_{cyc}x^2y^2=\left(\sum_{cyc}xy\right)^2-2xyz(x+y+z)=1$$ and $$x^2y^2z^2=1.$$
Thus, $$\sum_{cyc}x^8=\left(\sum_{cyc}x^4\right)^2-2\sum_{cyc}x^4y^4=$$
$$=\left(\left(\sum_{cyc}x^2\right)^2-2\sum_{cyc}x^2y^2\right)^2-2\left(\left(\sum_{cyc}x^2y^2\right)^2-2x^2y^2z^2\sum_{cyc}x^2\right)=$$
$$=(4-2)^2-2(1-2\cdot2)=10.$$
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