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

$$
\left\{\begin{array}{l}
x+y=1 \\
x^{3}+y^{3}=19
\end{array}\right.
$$
Once again, by trial and error, we find the solutions $(3,-2)$ and $(-2,3)$, and we will prove that these are the only ones.

We observe here that $(x+y)^{3}=\left(x^{3}+y^{3}\right)+3 x y(x+y)$, and therefore that

$$
x y=\frac{(x+y)^{3}-\left(x^{3}+y^{3}\right)}{3(x+y)}=\frac{1-19}{3}=-6
$$

We conclude that

$$
(t-x)(t-y)=t^{2}-t-6=(t-3)(t+2)
$$

and therefore that $(3,-2)$ and $(-2,3)$ are indeed the only solutions.