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 $a-b|P(a)-P(b)$ .

You can do this pretty systematically. The given information is equivalent to the statements that  $P_1(\alpha) - P_2(\alpha) = 0$ and  $P_2(\alpha) - P_3(\alpha) = 0$; in other words that
$$(a-b)\alpha^2 -(b - c)\alpha -(c - a) = 0$$
$$(b-c)\alpha^2 -(c - a)\alpha -(a - b) = 0$$
Next, it is natural to eliminate $\alpha^2$ and then solve for $\alpha$. So, we take $b-c$ times the first equation minus $a - b$ times the second equation to obtain
$$((a-b)(c-a) -(b-c)^2)\alpha + (a-b)^2 - (c-a)(b-c) = 0$$
We rewrite this as 
$$((a-b)(c-a) -(b-c)^2)\alpha =   (c-a)(b-c) - (a-b)^2$$
By symmetry we have also that
$$((b-c)(a-b) -(c-a)^2)\alpha =   (a-b)(c-a) - (b-c)^2$$
$$((c-a)(b-c) - (a-b)^2)\alpha =  (b-c)(a-b) -(c-a)^2$$
Adding the three equations gives
$$((a-b)(b-c) + (b-c)(a-b) + (c-a)(b-c) - ((a-b)^2 + (b-c)^2 + (c-a)^2))\alpha = ((a-b)(b-c) + (b-c)(a-b) + (c-a)(b-c) - ((a-b)^2 + (b-c)^2 + (c-a)^2))$$
For this to hold, either $\alpha = 1$ or 
$$(a-b)(b-c) + (b-c)(a-b) + (c-a)(b-c) - ((a-b)^2 + (b-c)^2 + (c-a)^2) = 0$$
Plugging in $\alpha = 1$ gives $a = b + c,$ $b = c+a,$ $c = a+ b$, leading immediately to $a = b = c$. One can use the AM-GM inequality on the cross terms to show that 
$$(a-b)(b-c) + (b-c)(a-b) + (c-a)(b-c) - ((a-b)^2 + (b-c)^2 + (c-a)^2) \leq  0$$
Equality holds iff $a-b = b-c = c-a$, which once again leads to $a = b = c$.
