# 1999 JBMO Problems/Problem 1

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## Problem

Let $a,b,c,x,y$ be five real numbers such that $a^3 + ax + y = 0$, $b^3 + bx + y = 0$ and $c^3 + cx + y = 0$. If $a,b,c$ are all distinct numbers prove that their sum is zero.

## Solution

After solving for $-y$ in all three equations, we have \begin{align*} -y &= a^3 + ax \\ &= b^3 + bx \\ &= c^3 + cx \end{align*} Thus, we know that $a^3 + ax = b^3 + bx = c^3 + cx$.

Since $a^3 + ax = b^3 + bx$, rearrange and factor terms to get \begin{align*} 0 &= a^3 + ax - b^3 - bx \\ &= (a-b)(a^2 + ab + b^2 + x) \end{align*} Since $a \ne b$, $a^2 + ab + b^2 = -x$. By using the same steps, $a^2 + ac + c^2 = -x$, so by substituting and rearranging terms, we have \begin{align*} a^2 + ac + c^2 &= a^2 + ab + b^2 \\ ac + c^2 &= ab + b^2 \\ -ab + ac &= b^2 - c^2 \\ -a(b-c) &= (b+c)(b-c) \\ 0 &= (a+b+c)(b-c) \end{align*} Since $b \ne c$, we must have $a+b+c = 0$.