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. 
 A: 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.
A: Given:
$$x^3+2016x+1=0 \tag1$$
$$y^3+2016x-1=0 \tag2$$. 
When you solve (1) for $x$:
You get a repeated root at $x=-0.000496032$.
When you solve equation (2) for $y$, you get $y=0.000496032$
We can use the values obtained to  calculate $x+y=0$
So the solutions are:
$y=0.000496032$, $x=-0.000496032$ or $x=0.000496032$, $y=-0.000496032$
Another approach is adding equations (1) and (2) to get:
$$x^3-y^3=(x-y)(x^2+xy+y^2)=0 \tag3$$
The expression $(x^2+xy+y^2)$ can't be zero for real values of x. This suggests that $x=y$. 
Substituting the value of $x$ in Equation (2) we get:
$$y^3+2016(-y)-1=0$$
which has a real solution $y=0.000496032$.
So the solutions are (for which $x+y=0$ and both $x,y$ are real):
$y=0.000496032$, $x=-0.000496032$ or $x=0.000496032$, $y=-0.000496032$
Note: I use the equal sign loosely here the accurate values may not be fully precise.
