TITLE: Polynomials Question: Proving $a=b=c$.
QUESTION [7 upvotes]: 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 $a-b|P(a)-P(b)$ .

REPLY [8 votes]: 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 $\alpha=1$
$a-b-c=b-c-a=c-a-b \implies a=b=c$.