Algebraic manipulation solving for $x+y$ There are real solutions $x,y$ to
$x^3+2016x+1=0$
$y^3+2016x-1=0$. 
Find $x+y$.
My thinking: adding them we get
$x^3+y^3+2016(x+y)=0$
$(x+y)(x^2-xy+y^2)+2016(x+y)=0$
$(x+y)(x^2-xy+y^2+2016)=0$
Then, either 
$x+y=0$ 
or, 
$x^2-xy+y^2+2016=0$. 
I’m not sure how to proceed with the second equation and the first seems a little odd. 

Note that 
$$x^2-xy+y^2+2016= \left( x-\frac{1}{2}y \right)^2 +\frac{3}{4}y^2 + 2016>0$$
So, the second equation 
$$x^2-xy+y^2+2016=0$$
has no real solutions. Thus, $x+y=0$ is the only solution.
