<s>[INST] 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.
 [/INST] Subtract consecutive equations to get$$a^3-b^3+(a-b)x=0\iff a^2+b^2+ab+x=0~~\because a\ne b\\b^3-c^3+(b-c)x=0\iff b^2+c^2+bc+x=0~~\because b\ne c$$
Subtract these equations to get$$a^2-c^2+b(a-c)=0\iff a+b+c=0~~\because a\ne c$$
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