Determine all real numbers $x, y, z$ satisfying $x+y+z=2, x^{2}+y^{2}+z^{2}=6$ and $x^{3}+y^{3}+z^{3}=8$.
We have the following system:

$$
\left\{\begin{array}{l}
s_{1}=2 \\
s_{1}^{2}-2 s_{2}=6 \\
s_{1}^{3}-3 s_{1} s_{2}+3 s_{3}=8
\end{array}\right.
$$

which is not hard to solve by

$$
\left\{\begin{array}{l}
s_{1}=2 \\
s_{2}=-1 \\
s_{3}=-2
\end{array}\right.
$$

$x, y, z$ are therefore roots of the polynomial $X^{3}-2X^{2}-X+2=(X^{2}-1)(X-2)=(X-1)(X+1)(X-2)$, so $x, y, z$ are a permutation of $-1,1,2$.