Berkeley Math Circle

Solution:
Let $u = a - b$, $v = b - c$, and $w = c - a$. Then $u + v + w = 0$, and we would like to prove that $u = v = w = 0$. We have
$$\begin{aligned} a x^{2} - b x - c &= b x^{2} - c x - a \\ (a - b) x^{2} + (-b + c) x + (-c + a) &= 0 \\ u x^{2} - v x - w &= 0. \end{aligned}$$

In a similar manner, we get
$$v x^{2} - w x - u = 0$$
and
$$w x^{2} - u x - v = 0.$$

Taking $v$ times (1) minus $u$ times (2) eliminates the $x^{2}$ term; we get
$$\left(v^{2} - w u\right) x = u^{2} - v w.$$

Letting $A = u^{2} - v w$, $B = v^{2} - w u$, $C = w^{2} - u v$, we now have $B x = A$, and analogously $C x = B$ and $A x = C$. We see that if any of $A$, $B$, and $C$ are zero, then they all are. So we have two cases:

Case 1. $A = B = C = 0$, that is, $u^{2} = v w$, $v^{2} = w u$, and $w^{2} = u v$. Clearly if one of $u$, $v$, and $w$ is $0$, then they all are and we are done. Otherwise, dividing the first equation by the second leads to $v^{3} = w^{3}$ and $v = w$. Similarly $u = v = w$. But since $u + v + w = 0$, the only common value that $u$, $v$, and $w$ can have is $0$ and we are done.

Case 2. $A$, $B$, and $C$ are all nonzero. Multiplying the three equations $B x = A$, $C x = B$, and $A x = C$ together, we derive that $x^{3} = 1$ so $x = 1$. Plugging $x = 1$ into the original equations gives
$$a - b - c = b - c - a = c - a - b,$$
so $a = b = c$ as desired.