11.3. The numbers $x, y$ satisfy the equation $\sqrt{x^{3}+y}+\sqrt{y^{3}+x}=\sqrt{x^{3}+x}+\sqrt{y^{3}+y}$. Can we assert that $x=y$?
Answer: it can be. Solution. See the solution to problem 10.3: everywhere instead of inequalities, equalities should be considered, and then the original equation reduces to the equation $(x-y)^{2}\left(x^{2}+x y+y^{2}\right)=0$. The first factor is zero (only) when $x=y$, and the second when $x=y=0$.