Let $x$ and $y$ be real numbers such that $x+y=10$ and $x^{3}+y^{3}=400$, determine the value of $x^{2}+y^{2}$.
Solution

Given

$$
\begin{aligned}
x^{3}+y^{3} & =(x+y)\left(x^{2}-x y+y^{2}\right) \\
& =(x+y)\left((x+y)^{2}-3 x y\right) \\
& =10(100-3 x y)
\end{aligned}
$$

it follows that $40=100-3 x y$, that is, $x y=20$. Therefore,

$$
\begin{aligned}
x^{2}+y^{2} & =(x+y)^{2}-2 x y \\
& =100-40 \\
& =60
\end{aligned}
$$