<s>[INST] Three polynomials which have the same value for a variable 
Let $P_1(x)= ax^2-bx-c$, $P_2(x)=bx^2-cx-a$, and $P_3(x)= cx^2-ax-b$ be three quadratic polynomials, where $a,b$, and $c$ are non-zero real numbers. Suppose that there exists a real number $k$ such that $$ P_1(k) = P_2(k)= P_3(k) $$ Prove that $a=b=c$.

Hello everybody! The above is a question I got stuck on. This problem is from an Indian Olympiad. Here is what I did: 

After manipulating the given information, we derive at $(a-b)k^2-(b-c)k-(c-a)=0$ $(b-c)k^2-(c-a)k-(a-b)=0 $ and $(c-a)k^2 - (a-b)k - (b-c)$. Now the solution involves some addition and subtraction to factor it and get $a=b=c$.

But my question is: Can’t I just compare the coefficients of the three equations because they are all zero from which we get $b-c = c-a$, $c-a = a-b $ and $a-b = b-c$ and get $a=b=c$? If not, why?
Thanks
 [/INST] Write $P_1(k) = P_2(k)= P_3(k)=v$. Then 
$$
\pmatrix{ k^2 & -k & -1 \\ -1 & k^2 & -k \\ -k & -1 & k^2}
\pmatrix{ a \\ b \\ c}
=
\pmatrix{ v \\ v \\ v}
$$
and so
$$
\pmatrix{ a \\ b \\ c}
=
\frac{1}{k^6 - 4 k^3 - 1}
\pmatrix{k^4 - k & k^3 + 1 & 2 k^2 \\
2 k^2 & k^4 - k & k^3 + 1 \\
k^3 + 1 & 2 k^2 & k^4 - k}
\pmatrix{ v \\ v \\ v}
$$
Since the sums of the elements in each row are the same, we get $a=b=c$.
Alternatively, without any computation, just note that the first matrix is a circulant matrix and so has a circulant inverse.
On the other hand, we need $k^6 - 4 k^3 - 1 \ne 0$, that is, $k \ne (1\pm\sqrt5)/2$, the roots of $k^2-k-1$. This case needs to be handled separately but it still works. A necessary condition is $v=0$.
(Actually, $a=b=c\ne0$ implies $k^2-k-1=0$.)
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