From the system of equations prove $a+b+c=0$ 
$a,b,c$ are distinct real numbers and $x,y$ are also real numbers. we have these equations:
$${ \begin{cases}{a^3+ax+y=0} \\ {b^3+bx+y=0} \\
 {c^3+cx+y=0}\end{cases} }$$ Prove $a+b+c=0$

I added all the equations together and get:
$$a^3+b^3+c^3+(a+b+c)x+3y=0$$
It is similar to Euler identity (because we have $a^3+b^3+c^3$). if $a+b+c=0$ then from Euler identity we can conclude $a^3+b^3+c^3=3abc$. and equation change to:
$$3abc+3y=0$$
But it seems that doesn't work.

Here's an alternative solution to Albus Dumbledore's. Note that the three equations imply that the nonzero vector $(1,x,y)$ is in the kernel of
$$\begin{pmatrix}a^3&a&1\\b^3&b&1\\c^3&c&1\end{pmatrix}$$
of determinant (after factoring)
$$(a-b)(a-c)(b-c)(a+b+c).$$
But also, the determinant must be 0 since it has nontrivial kernel. So $(a-b)(a-c)(b-c)(a+b+c)=0$ and by the assumption that $a,b,c$ are distinct, the result follows.
