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.
 A: 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 $\ge0\implies \cdots$
A: Let us put $w=P_1(v)=P_2(v)=P_3(v)$. We have 
$$
\begin{array}{lcl}
P_1(v)=av^2-bv-c, & -(v+v^2)P_1(v) &=& a(-v^4-v^3)+b(v^3+v^2)+c(v^2+v) \\
P_2(v)=bv^2-cv-a, & (v^2+1)P_2(v) &=& a(-v^2-1)+b(v^4+v^2)+c(-v^3-v) \\
P_3(v)=cv^2-av-b, & (v-1)P_3(v) &=& a(-v^2+v)+b(-v+1)+c(v^3-v^2) \\
\end{array}
$$
So if we put $z_1=-(v+v^2)P_1(v)+(v^2+1)P_2(v)+(v-1)P_3(v)$, we have
$z_1=-(v+v^2)w+(v^2+1)w+(v-1)w=0$ but on the other hand
$z_1=(b-a)(v^4+v^3+2v^2-v+1)$. So
$$
(v^4+v^3+2v^2-v+1)(b-a)=0 \tag{1}
$$
Similarly, 
$$
\begin{array}{lcl}
P_1(v)=av^2-bv-c, & -(v^2+1)P_1(v) &=& a(-v^4-v^2)+b(v^3+v)+c(v^2+1) \\
P_2(v)=bv^2-cv-a, & (-v+1)P_2(v) &=& a(v-1)+b(-v^3+v^2)+c(v^2-v) \\
P_3(v)=cv^2-av-b, & (v^2+v)P_3(v) &=& a(-v^3-v^2)+b(-v^2-v)+c(v^4+v^3) \\
\end{array}
$$
So if we put $z_2=-(v^2+1)P_1(v)+(-v+1)P_2(v)+(v^2+v)P_3(v)$, we have
$z_2=-(v^2+1)w+(-v+1)w+(v^2+v)w=0$ but on the other hand
$z_2=(c-a)(v^4+v^3+2v^2-v+1)$. So
$$
(v^4+v^3+2v^2-v+1)(c-a)=0 \tag{2}
$$
In view of (1) and (2), it will suffice to show that $u=v^4+v^3+2v^2-v+1$ is nonzero.
But 
$$
u=\frac{15}{4}v^2+(2v^2+v-2)^2 \tag{3}
$$
so we are done.
