Find the triplets of real numbers $x, y, z$ satisfying:

$$
\begin{aligned}
x+y+z & =2 \\
x^{2}+y^{2}+z^{2} & =26 \\
x^{3}+y^{3}+z^{3} & =38
\end{aligned}
$$

## Snacking
We use elementary symmetric polynomials: let $\sigma_{1}=x+y+z$, $\sigma_{2}=xy+yz+zx$, and $\sigma_{3}=xyz$. We know (see course) that $x, y, z$ are the roots of the polynomial $P(X)=X^{3}-\sigma_{1} X^{2}+\sigma_{2} X-\sigma_{3}$.

We have directly $\sigma_{1}=2$. And, $4=(x+y+z)^{2}=x^{2}+y^{2}+z^{2}+2 \sigma_{2}=26+2 \sigma_{2}$, so $\sigma_{2}=-11$. For $\sigma_{3}$, which is of degree 3, we look for an expression of degree 3 involving things we already know. Let's try $(x+y+z)\left(x^{2}+y^{2}+z^{2}\right)$ (which is a bit simpler than $(x+y+z)^{3}$):

$$
\begin{aligned}
& 52=x^{3}+y^{3}+z^{3}+x^{2} y+y^{2} x+x^{2} z+z^{2} x+y^{2} z+z^{2} y \\
& 14=x^{2} y+y^{2} x+x^{2} z+z^{2} x+y^{2} z+z^{2} y
\end{aligned}
$$

This is not enough, so let's use $(x+y+z)^{3}$:

$$
\begin{aligned}
8 & =x^{3}+y^{3}+z^{3}+3\left(x^{2} y+y^{2} x+x^{2} z+z^{2} x+y^{2} z+z^{2} y\right)+6 \sigma_{3} \\
8 & =38+42+6 \sigma_{3} \\
\sigma_{3} & =-12
\end{aligned}
$$

Thus, $x, y, z$ are the roots of $P(X)=X^{3}-2 X^{2}-11 X+12$. We notice that 1 is an obvious solution. Factoring, $P(X)=(X-1)\left(X^{2}-X-12\right)$. We solve the quadratic trinomial and find $P(X)=(X-1)(X+3)(X-4)$. Therefore, we have $x=1, y=-3$, and $z=4$, up to permutation.