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.
 A: 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.
A: 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 $10$.
A: Denoting with $e_k(x,y,z), k\geq 0$ the elementary symmetric polynomials
\begin{align*}
e_1(x,y,z)&=x+y+z=0\\
e_2(x,y,z)&=xy+xz+yz=-1\tag{1}\\
e_3(x,y,z)&=xyz=-1
\end{align*}
and with $p_k(x,y,z), k\geq 0$ the $k$-th power sum
\begin{align*}
p_k(x,y,z)=x^k+y^k+z^k\tag{2}
\end{align*}
we recall Newtons identities
admit a generating function representation of the power sums as
\begin{align*}
\color{blue}{\sum_{k=1}^\infty(-1)^{k-1}p_k\frac{t^k}{k}=\ln(1+e_1t+e_2t^2+e_3t^3+\cdots)}\tag{3}
\end{align*}

We obtain from (1) - (3)
\begin{align*}
\color{blue}{\left.x^8+y^8+z^8\right|_{{{e_1=0\ \ }\atop{e_2=-1}}\atop{e_3=-1}}}
&=(-8)[t^8]\ln\left(1-t^2-t^3\right)\\
&=(-8)[t^8]\ln\left(1-t^2(1+t)\right)\\
&=8[t^8]\sum_{j=1}^\infty\frac{t^{2j}(1+t)^j}{j}\tag{4.1}\\
&=8\sum_{j=1}^{4}\frac{1}{j}[t^{8-2j}](1+t)^j\tag{4.2}\\
&=8\sum_{j=1}^4\frac{1}{j}\binom{j}{8-2j}\tag{4.3}\\
&=8\left(\frac{1}{3}\binom{3}{2}+\frac{1}{4}\binom{4}{0}\right)\tag{4.4}\\
&\,\,\color{blue}{=10}
\end{align*}

Comment:

*

*In (4.1) we expand the logarithmic series.


*In (4.2) we apply the rule $[t^{p-q}]A(t)=[t^p]t^qA(t)$. We also set the upper limit of the series to $4$ since other terms do not contribute.


*In (4.3) we select the coefficient of $t^{8-2j}$.


*In (4.4) we use $\binom{p}{q}=0$ if $0<p<q$ are positive integers.
A: $x, y$ and $z$ are roots of the polynomial $x^3-x+1=0$
In the usual notation ,we have $$\Sigma x^2=(\Sigma x)^2-2\Sigma xy=0-2(-1)=2$$
Then since the relation $x^3=x-1$ holds for all three roots, summing these gives:
$$\Sigma x^3=\Sigma x - \Sigma 1=0-3=-3$$
Likewise,
$$\Sigma x^4=\Sigma x^2-\Sigma x=2$$
$$\Sigma x^5=\Sigma x^3-\Sigma x^2=-3-2=-5$$
$$\Sigma x^6=\Sigma x^4-\Sigma x^3=2--3=5$$
And finally,
$$\Sigma x^8=\Sigma x^6-\Sigma x^5=5--5=10$$
A: 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 $x+y+z=0$:
$$
\require{cancel}
\begin{align}
x^8+y^8+z^8 &= 2(x^2+y^2+z^2) - \cancel{3(x+y+z)} + 6
\\ &= 2\left(\bcancel{(x+y+z)^2}-2(xy+yz+zx)\right)+6
\\ &= -4 \cdot (-1) + 6
\\ &= 10
\end{align}
$$
A: 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,
$$x^8 + y^8 + z^8 = -(-10) = 10.$$
A: Question: "@hm2020 ,hi , I can't relate the equation you are proposing to the problem, could you clarify? – Nabla"
Answer: There is an explicit formula (you find it on wikipedia under the name "the Cardano formula", or in algebra books) for the roots of $f(t):=t^3-t+1$: Using the formula it follows the following number is a root:
$$x:=\sqrt[3]{-\frac{1}{2}+\sqrt{\frac{23}{108}}}+\sqrt[3]{-\frac{1}{2}-\sqrt{\frac{23}{108}}   }$$
It follows you may (by hand, in principle) use polynomial division to calculate a decomposition $f(t)=(t-x)p(t)$ where $p(t)$ is a degree 2 polynomial and there is an explicit formula for the roots of $p(t)$. Hence you may give explicit formulas for all the roots $x,y,z$ of $f(t)$ and also calculate all powers $x^m+y^n+z^k$ for all integers $m,n,k$. This is a "brute force" method but it still gives explicit formulas. You get the following:
$$f(t)=(t-x)(t^3+xt+(x^2-1))=(t-x)p(t)$$
where $p(t)=t^2+xt+(x^2-1)$. The polynomial $p(t)$ has roots
$$y:= \frac{1}{2}(-x +\sqrt{4-3x^2})\text{  and } z:=\frac{1}{2}(-x-\sqrt{4-3x^2}).$$
The above formulas give all roots $x,y,z$ of $f(t)$ and you get
$$F1\text{  }x^m+y^n+z^k:= (\sqrt[3]{-\frac{1}{2}+\sqrt{\frac{23}{108}}}+\sqrt[3]{-\frac{1}{2}-\sqrt{\frac{23}{108}}   })^m+$$
$$(\frac{1}{2}(-x +\sqrt{4-3x^2}))^n+(\frac{1}{2}(-x-\sqrt{4-3x^2}))^k.$$
The formula $F1$ gives an explicit formula for all $m,n,k$, in particular for $m=n=k=8$.
You must check the "Cardano formula": It says that a solution to $g(t):=t^3+pt+q=0$  is the following:
$$x:=\sqrt[3]{-\frac{q}{2}+\sqrt{\frac{q^2}{4}+\frac{p^3}{27}}}+\sqrt[3]{-\frac{q}{2}-\sqrt{\frac{q^2}{4}+\frac{p^3}{27}}   }.$$
The above formulas give all solutions to the equation $g(t)=0$. The above calculation give explicit formulas for numbers $x(p,q), y(p,q), z(p,q)$ depending on $p,q$ solving $g(t)=0$.
A: 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, $$\sum_{cyc}x^8=\left(\sum_{cyc}x^4\right)^2-2\sum_{cyc}x^4y^4=$$
$$=\left(\left(\sum_{cyc}x^2\right)^2-2\sum_{cyc}x^2y^2\right)^2-2\left(\left(\sum_{cyc}x^2y^2\right)^2-2x^2y^2z^2\sum_{cyc}x^2\right)=$$
$$=(4-2)^2-2(1-2\cdot2)=10.$$
