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.

Note that $x$, $y$, $z$ are the roots of the polynomial $\lambda^3 - \lambda + 1$, so they are the  eigenvalues of  the companion matrix
$$A=\left(\begin{matrix}
0 & 0 & -1 \\
1 & 0 & 1\\
0 & 1 & 0
\end{matrix} \right)$$
Conclude that the eigenvalues $A^{8}$ are $x^8$, $y^8$, $y^8$. Calculate $A^8$ by squaring three times ( @Theo Bendit:'s idea in the comments). We get
$$A^8=\left(\begin{matrix}
2 & -2 & 3 \\
-3 & 4 & -5\\
2 & -3 & 4
\end{matrix} \right)$$
so the sum $x^8+y^8+z^8$ equals the trace of $A^8$, that is $10$.
