<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] This is an expansion of my comment. Recall Vieta's formulas (in this case, for cubics):

If $p, q, r$ are the roots of $t^3 + bt^2 + ct + d$, where $a \neq 0$, then
\begin{align*}
p + q + r &= -b \\
pq + qr + pr &= c \\
pqr &= -d.
\end{align*}

This follows easily by just expanding $(t - p)(t - q)(t - r)$ and equating coefficients with $t^3 + bt^2 + ct + d$.
Using these formulas, we can see, as others have commented, that the numbers $x, y, z$ in your question are just roots of the polynomial $t^3 - t + 1$. We can also use Vieta's formulas, and a little computation along the lines of what you've attempted already, to obtain a polynomial whose roots are precisely the squares of the roots of the original polynomial.
Let's say that $t^3 + bt^2 + ct + d$ has roots $p, q, r$ as above. Let $t^3 + b't^2 + c't + d'$ have roots $p^2, q^2, r^2$. We wish to compute $b', c' d'$ vie Vieta's formulas, i.e. to compute $p^2 + q^2 + r^2$, $p^2 q^2 + q^2 r^2 + p^2 r^2$, and $p^2 q^2 r^2$ in terms of $a, b, c$.
There's a quick and easy one: $p^2 q^2 r^2 = (pqr)^2 = d^2$. Thus $d' = -d^2$.
To compute $p^2 + q^2 + r^2$, consider
$$b^2 = (p + q + r)^2 = p^2 + q^2 + r^2 + 2(pq + qr + rp) = -b' + 2c,$$
thus $b' = 2c - b^2$.
Lastly, we compute
\begin{align*}
c^2 &= (pq + qr + pr)^2 \\
&= p^2 q^2 + q^2 r^2 + p^2r^2 + 2p^2qr + 2pq^2r + 2pqr^2 \\
&= c' + 2pqr(p + q + r) \\
&= c' + 2bd,
\end{align*}
so $c' = c^2 - 2bd$.
Thus, our transformation which squares roots takes a monic polynomial $t^3 + bt^2 + ct + d$ and transforms it into the monic polynomial
$$t^3 + (2c - b^2)t^2 + (c^2 - 2bd)t - d^2.$$
Let's apply it to $t^3 - t + 1$ to get a polynomial whose roots are $x^2, y^2, z^2$. We obtain the polynomial:
$$t^3 + (2(-1) - 0^2)t^2 + ((-1)^2 - 2(0)(1))t - 1^2 = t^3 - 2t^2 + t - 1.$$
Apply again to get a polynomial whose roots are $x^4, y^4, z^4$:
$$t^3 + (2(1) - (-2)^2)t^2 + (1^2 - 2(-2)(-1))t - (-1)^2 = t^3 - 2t^2 - 3t - 1.$$
We could apply this transformation one final time to get the polynomial whose roots are $x^8, y^8, z^8$, but really we just need the coefficient of $t^2$. So, partially applying the above formula, we get:
$$t^3 + (2(-3) - (-2)^2)t^2 + \underline{\hspace{12pt}} t + \underline{\hspace{12pt}} = t^3 - 10t^2 + \underline{\hspace{12pt}} t + \underline{\hspace{12pt}}.$$
So, by Vieta,
$$x^8 + y^8 + z^8 = -(-10) = 10.$$
</s>