Polynomials Question: Proving $a=b=c$. Question:

Let $P_1(x)=ax^2-bx-c, P_2(x)=bx^2-cx-a \text{   and } P_3=cx^2-ax-b$
  , where $a,b,c$ are non zero reals. There exists a real $\alpha$ such
  that $P_1(\alpha)=P_2(\alpha)=P_3(\alpha)$. Prove that $a=b=c$.

The questions seems pretty easy for people who know some kind of calculus. Since, the question is from a contest, no calculus can be used. I have got a solution which is bit long(No calculus involved), I'm looking for a simplest way to solve this, which uses more direct things like $a-b|P(a)-P(b)$ .

Observe that
$$ P_1 ^2 + P_2^2 + P_3^2 - P_1P_2 - P_2 P_3 - P_3 P_1 = (a^2+b^2+c^2-ab-bc-ca) ( x^4 +x^3 + x^2 + x + 1) $$
This has a real zero. But since $x^4 + x^3 + x^2 + x + 1 = 0 $ has no real roots, thus the coefficient must be 0.
Hence, $0 = a^2+b^2+c^2-ab-bc-ca = \frac{1}{2} [(a-b)^2 + (b-c)^2 + (c-a)^2] $, so $a=b=c$.

The idea of the first equation comes from trying to exploit the symmetry in $P_1 - P_2 = (a-b) x^2 - (b-c) x - (c-a)$, but being stumped by the negative signs. Taking the squares made the most sense, and the sum gives you the symmetry of the RHS.
