Find all triplets $x, y, z$ such that

\[
\begin{aligned}
x+y+z & =5 \\
x^{2}+y^{2}+z^{2} & =19 \\
x^{3}+y^{3}+z^{3} & =53
\end{aligned}
\]
We once again use the recurrence formula for symmetric polynomials: here $S_{1}=5, S_{2}=19$ and $S_{3}=53$, so $\sigma_{1}=S_{1}=5, \sigma_{2}=\frac{1}{2}\left(\sigma_{1} S_{1}-S_{2}\right)=3$ and $\sigma_{3}=\frac{1}{3}\left(S_{3}-\sigma_{1} S_{2}+\sigma_{2} S_{1}\right)=-9$. The real numbers $x, y, z$ we are looking for are therefore the roots of the polynomial $X^{3}-\sigma_{1} X^{2}+\sigma_{2} X-\sigma_{3}=X^{3}-5 X^{2}+3 X+9=(-3+X)^{2}(1+X)$. Up to the symmetry of the variables $x, y$ and $z$, we were therefore looking for the triplet $(-1, \sqrt{3} i,-\sqrt{3} i)$.