Find all real numbers $(x, y, z)$ such that

$$
\left\{\begin{array}{c}
x+y+z=2 \\
x^{2}+y^{2}+z^{2}=6 \\
x^{3}+y^{3}+z^{3}=8
\end{array}\right.
$$
As in the previous exercise, we seek the remaining Viète relations: $x y+y z+z x$ and $x y z$. $(x+y+z)^{2}=x^{2}+y^{2}+z^{2}+2(x y+y z+z x) \Longleftrightarrow x y+y z+z x=-1$. Furthermore, $(x+y+z)^{3}=x^{3}+y^{3}+z^{3}+3\left(x y^{2}+x z^{2}+y x^{2}+y z^{2}+z x^{2}+z y^{2}+2 x y z\right)=x^{3}+y^{3}+z^{3}+$ $3((x y+y z+z x)(x+y+z)-x y z) \Longleftrightarrow x y z=(x y+y z+z x)(x+y+z)=-2$.

Thus, $x, y, z$ are solutions of $X^{3}-2 X^{2}-X-2$. We notice that 1 is a root. $X^{3}-2 X^{2}-$ $X-2=(X-1)\left(X^{2}-3 X+2\right)=(X-1)(X+1)(X-2)$ Thus, $(x, y, z)=(-1,1,2)$ and all its permutations.