8.4. It is known that for real numbers $a$ and $b$ the following equalities hold:

$$
a^{3}-3 a b^{2}=11, \quad b^{3}-3 a^{2} b=2
$$

What values can the expression $a^{2}+b^{2}$ take?
Answer: 5.

Solution. We have

$$
\left(a^{2}+b^{2}\right)^{3}=a^{6}+3 a^{4} b^{2}+3 a^{2} b^{4}+b^{6}=\left(a^{3}-3 a b^{2}\right)^{2}+\left(b^{3}-3 a^{2} b\right)^{2}=11^{2}+2^{2}=125
$$

From this, $a^{2}+b^{2}=5$.

Comment. A correct and justified solution - 7 points. The correct answer obtained based on a simple example satisfying the condition ( $a=-1, b=2$ ), but not proven that the same answer for other pairs - 3 points. Some progress - 2 points. The solution is started, but the progress is insignificant - 1 point. Only the answer is provided - 0 points. The problem is not solved or solved incorrectly - 0 points.