Find all real $(x, y)$ such that $x+y=1$ and $x^{3}+y^{3}=19$
We are looking to calculate $xy$, because according to Viète's formulas, we can then find $x, y.1=$ $(x+y)^{3}=x^{3}+y^{3}+3xy(x+y) \Longleftrightarrow 1=19+3xy \Longleftrightarrow xy=-6$. Thus $x, y$ are the roots of $X^{2}-X-6$. After a bit of trial and error, we find -2 and 3 as roots (we can also use the discriminant method in more complex cases). Therefore, $(x, y)=(-2,3)$ or $(x, y)=(3,-2)$