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.

The following is an elementary, entirely self-contained solution which does not assume knowledge of Vieta's relations, Newton's identities, or anything other than simple algebra. (This is not the best or recommended way to solve it - see the other answers and comments for that - but it answers OP's edit asking for a more basic solution.)

*

*From $x+y+z=0$ it follows that:

$$y+z = -x \tag{1}$$

*

*From $xy +xz+yz=-1$ and $(1)$ it follows that:
$$-1=x(y+z)+yz=-x^2 + yz \;\;\implies\;\; yz = x^2-1 \tag{2}$$


*From $xyz=-1$ and $(2)$ it follows that:
$$-1=xyz=x(x^2-1)=x^3-x \;\;\implies\;\; x^3=x-1 \tag{3}$$
Multiplying $(3)$ by $x$ gives:
$$x^4=x^2-x \tag{4}$$
Squaring $(4)$ and using that $x^3=x-1$ per $(3)\,$:
$$
\begin{align}
x^8 &= x^4-2x^3+x^2
\\ &=x \cdot x^3-2x^3+x^2
\\ &=x(x-1)-2(x-1)+x^2
\\ &=2x^2-3x+2 \tag{5}
\end{align}
$$
Repeating the steps $(1)\dots(5)$ for the other variables, it follows by symmetry that:
$$
\begin{align}
y^8 = 2y^2-3y+2 \tag{6}
\\ z^8 = 2z^2-3z+2 \tag{7}
\end{align}
$$
Adding $(5)+(6)+(7)\,$ and using that $x+y+z=0$:
$$
\require{cancel}
\begin{align}
x^8+y^8+z^8 &= 2(x^2+y^2+z^2) - \cancel{3(x+y+z)} + 6
\\ &= 2\left(\bcancel{(x+y+z)^2}-2(xy+yz+zx)\right)+6
\\ &= -4 \cdot (-1) + 6
\\ &= 10
\end{align}
$$
