Solve the equation $\left(x^{2}+y^{2}\right)^{3}=\left(x^{3}-y^{3}\right)^{2}$ on the set of real numbers.
Solution. By expanding the parentheses, rearranging the equation to zero, and performing factorizations, we obtain:

$$
\begin{aligned}
x^{6}+y^{6}+3 x^{4} y^{2}+3 x^{2} y^{4} & =x^{6}+y^{6}-2 x^{3} y^{3} \\
3 x^{4} y^{2}+3 x^{2} y^{4}+2 x^{3} y^{3} & =0 \\
x^{2} y^{2}\left(3 x^{2}+3 y^{2}+2 x y\right) & =0 \\
x^{2} y^{2}\left((x+y)^{2}+2\left(x^{2}+y^{2}\right)\right) & =0
\end{aligned}
$$

A product is zero if and only if one of its factors is zero. Three cases are possible:

1. $x^{2}=0$, so $x=0$, in which case $y$ can be any value.
2. $y^{2}=0$, so $y=0$, in which case $x$ can be any value.
3. $(x+y)^{2}+2 x^{2}+2 y^{2}=0$. The expression is the sum of three squares. Every square is non-negative, so the sum can only be zero if each term is zero. Therefore, $2 x^{2}=0$, so $x=0$; $2 y^{2}=0$, so $y=0$. In this case, $x+y=0$ is also true, so $(x+y)^{2}=0$ is satisfied. However, this solution is already included in cases 1 and 2.

The solutions are: $x=0$ and $y$ is any real number; $y=0$ and $x$ is any real number.