7. (4 points) Find what $x+y$ can be equal to, given that $x^{3}+6 x^{2}+16 x=-15$ and $y^{3}+6 y^{2}+16 y=-17$.
Answer: -4

Solution: Let $u=x+2$ and $v=y+2$. Then the original equations transform into $u^{3}+4 u=1$ and $v^{3}+4 v=-1$. Adding these equations, we get $(u+v)\left(u^{2}-u v+v^{2}+4\right)=0$. The second bracket is always positive, so the first one must be zero, from which we have $x+y=u+v-4=-4$.