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 $x+y+z=0$:

$$ \require{cancel} \begin{align} x^8+y^8+z^8 &= 2(x^2+y^2+z^2) - \cancel{3(x+y+z)} + 6 \\ &= 2\left(\bcancel{(x+y+z)^2}-2(xy+yz+zx)\right)+6 \\ &= -4 \cdot (-1) + 6 \\ &= 10 \end{align} $$

If you're not familiar to Girard-Newton identities, a recurrence relation may help:

\begin{align} \qquad a^n+b^n+c^n & = (a+b+c)(a^{n-1}+b^{n-1}+c^{n-1}) \\ & \quad-(bc+ca+ab)(a^{n-2}+b^{n-2}+c^{n-2})+abc(a^{n-3}+b^{n-3}+c^{n-3}) \end{align}

By successive iterations, you can raise up to the desired order.

Using SymmetricReduction[x^8+y^8+z^8,{x,y,z}] in Wolfram Alpha, the result is here.