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 $a-b|P(a)-P(b)$ .
 A: We have,
$a \alpha^2-b \alpha-c= b \alpha^2-c \alpha-a =c \alpha ^2-a \alpha-b=k$
Eliminating $\alpha^2$ would give me,
$(ca-b^2)\alpha-(bc-a^2)=k(b-a)$
$(ab-c^2) \alpha-(ca-b^2) =k(c-b)$
$(bc-a^2)\alpha -(ab-c^2)=k(a-c)$
Adding all of them would give me,
$(a^2+b^2+c^2-ab-ca-bc)(\alpha-1)=0 \implies \alpha =1$ or the other term is zero.
We know what happens when  $(a^2+b^2+c^2-ab-ca-bc)$ is zero. When $\alpha=1$
$a-b-c=b-c-a=c-a-b \implies a=b=c$.
A: Hint: if $a=b=c$ then all three polynomials are equal.  A useful trick to show that polynomials are equal is the following: if a polynomial $Q$ of degree $n$ (like $P_1-P_2$) has $n+1$ distinct roots (points $\beta$ such that $Q(\beta)=0$) then $Q$ is the zero polynomial.  In particular, if a quadratic has three zeroes, then it must be identically zero.  It follows that any two quadratics which agree at three distinct points must be identical.  (So you should try to construct a quadratic from $P_1,P_2,P_3$ that has three distinct zeroes and somehow conclude from that that $a=b=c$.)  
This result can be proved using the factor theorem, and requires no calculus (indeed, like the result of your question, it holds in polynomial rings where analysis can't be developed in a particularly meaningful sense, so any proof using calculus is rather unsatisfactory).  
Disclaimer: I haven't actually checked to see whether this approach works, but it's more or less the only fully general trick to show that two polynomials are the same.  
A: 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.
A: 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 $a = b = c$.
