<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] Hint: $a,b,c$ are the roots of cubic $$p(t)=t^3+tx+y$$

so by vieta $a+b+c=0$

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