Find the real numbers $x, y$ and $z$ such that

$$
\left\{\begin{array}{l}
x+y+z=3 \\
x^{2}+y^{2}+z^{2}=3 \\
x^{3}+y^{3}+z^{3}=3
\end{array}\right.
$$
First, by trial and error, we find that $x=y=z=1$ works. We will now prove that this is the only solution. To do this, 1 the idea is to obtain several polynomials of the degrees that interest us, in order to get expressions that we can use.

For example, $(x+y+z)^{2}=\left(x^{2}+y^{2}+z^{2}\right)+2(x y+y z+z x)$, so

$$
x y+y z+z x=\frac{(x+y+z)^{2}-\left(x^{2}+y^{2}+z^{2}\right)}{2}=\frac{3^{2}-3}{2}=3
$$

Similarly,

$$
\begin{aligned}
& (x+y+z)^{3}=\left(x^{3}+y^{3}+z^{3}\right)+3\left(x^{2} y+x^{2} z+y^{2} x+y^{2} z+z^{2} x+z^{2} y\right)+6 x y z \text { and } \\
& (x+y+z)\left(x^{2}+y^{2}+z^{2}\right)=\left(x^{3}+y^{3}+z^{3}\right)+\left(x^{2} y+x^{2} z+y^{2} x+y^{2} z+z^{2} x+z^{2} y\right) .
\end{aligned}
$$

We deduce that

$$
\begin{aligned}
x y z & =\frac{(x+y+z)^{3}-\left(x^{3}+y^{3}+z^{3}\right)-3(x+y+z)\left(x^{2}+y^{2}+z^{2}\right)+3\left(x^{3}+y^{3}+z^{3}\right)}{6} \\
& =\frac{3^{3}-3-3^{3}+3^{2}}{6} \\
& =1
\end{aligned}
$$

We conclude that

$$
(t-x)(t-y)(t-z)=t^{3}-3 t^{2}+3 t-1=(t-1)^{3}
$$

and therefore that $x=y=z=1$.