Prove $a = b = c$, given $P_1(x) = ax^2-bx-c$ , $P_2(x) = bx^2-cx-a$, $P_3(x)=cx^2-ax-b$ and $P_1(v)=P_2(v)=P_3(v)$ Prove $a = b = c$, given $P_1(x) = ax^2-bx-c$, $P_2(x) = bx^2-cx-a$, $P_3(x)=cx^2-ax-b$ and $P_1(v)=P_2(v)=P_3(v)$
where $v$ is a real number.
$a,b,c$ are non zero real numbers.

Let $P_1(v)=P_2(v)=P_3(v)=k$
So, $\displaystyle av^2-bv-c=k\iff av^2-bv-k=c\ \ \ \ (1)$  
$bv^2-cv-a=k\iff bv^2-cv-k=a \ \ \ \ (2)$
$cv^2-av-b=k\iff cv^2-av-k=b\ \ \ \ (3)$
Applying Cramer's Rule for $v^2,v,k$ we get $$\frac{v^2}{\sum_{\text{cyc}}(a^2-bc)}=\frac v{-\sum_{\text{cyc}}(a^2-bc)}=\frac k{-(a+b+c)\sum_{\text{cyc}}(a^2-bc)}=\frac1{-\sum_{\text{cyc}}(a^2-bc)}$$
If $\displaystyle\sum_{\text{cyc}}(a^2-bc)\ne0, v^2=-1$ and $v=1$ which is clearly impossible
$\displaystyle\implies \sum_{\text{cyc}}(a^2-bc)=0\implies(a-b)^2+(b-c)^2+(c-a)^2=0$
As $a,b,c$ are real, each addend must be $\ge0\implies \cdots$
