5. Let $a$ and $b$ be real numbers such that

$$
a^{3}-3 a b^{2}=44 \text{ and } b^{3}-3 a^{2} b=8
$$

Determine $a^{2}+b^{2}$.
Solution. We have:

$$
\begin{aligned}
\left(a^{2}+b^{2}\right)^{3} & =a^{6}+3 a^{4} b^{2}+3 a^{2} b^{4}+b^{6} \\
& =a^{6}-6 a^{4} b^{2}+9 a^{2} b^{4}+b^{6}-6 a^{2} b^{4}+9 a^{4} b^{2} \\
& =\left(a^{3}-3 a b^{2}\right)^{2}+\left(b^{3}-3 a^{2} b\right)^{2} \\
& =44^{2}+8^{2}=2000
\end{aligned}
$$

so therefore

$$
a^{2}+b^{2}=\sqrt[3]{2000}=10 \sqrt[3]{2}
$$