Folklore

Find $x^{3}+y^{3}$, given that $x+y=5$ and $x+y+x^{2} y+x y^{2}=24$.

#
Since $x+y+x^{2} y+x y^{2}=x+y+x y(x+y)=(x+y)(x y+1)=24$, using the condition $x+y=5$, we get $x y=3.8$.

We can proceed in different ways:

First method. $x^{3}+y^{3}(x+y)^{3}-3 x y(x+y)=125-3 \times 3.8 \times 5=68$.

Second method. $x^{3}+y^{3}=(x+y)\left(x^{2}-x y+y^{2}\right)=(x+y)\left((x+y)^{2}-3 x y\right)=5(25-3 \times 3.8)=68$.

## Answer

68.