7. (4 points) Find what $x+y$ can be equal to, given that $x^{3}-6 x^{2}+15 x=12$ and $y^{3}-6 y^{2}+15 y=16$.
Answer: 4.

Solution:

Let $u=x-2$ and $v=y-2$. Then the original equations transform into $u^{3}+3 u=-2$ and $v^{3}+3 v=2$. Adding these equations, we get $(u+v)\left(u^{2}-u v+v^{2}+3\right)=0$. The second bracket is always positive, so the first one must be zero, from which we have $x+y=u+v+4=4$.