Is value of $\alpha$ defined? Consider three quadratic functions:
$$P_1(x)=ax^2-bx-c$$
$$P_2(x)=bx^2-cx-a$$
$$P_3(x)=cx^2-ax-b$$ Where $a,b,c \in \mathbb{R}\backslash \left\{0\right\}$
If there exists real number $\alpha$ such that
$$P_1(\alpha)=P_2(\alpha)=P_3(\alpha)$$
Prove that $a=b=c$
My Try:
Using $$P_1(\alpha)=P_2(\alpha)=P_3(\alpha)$$
 we get
$$(a-b)\alpha^2-(b-c)\alpha-(c-a)=0 \tag{1}$$
$$(b-c)\alpha^2-(c-a)\alpha-(a-b)=0 \tag{2}$$
$$(c-a)\alpha^2-(a-b)\alpha-(b-c)=0 \tag{3}$$
Letting $p=a-b$,$q=b-c$ and $r=c-a$ we get the above three equations as:
$$p\alpha^2-q\alpha-r=0$$
$$q\alpha^2-r\alpha-p=0$$
$$r\alpha^2-p\alpha-q=0$$
Using $p+q+r=0$ and Eliminating $\alpha$ from last two equations above we get:
$$\alpha=\frac{p^2+pr+r^2}{p^2+pr+r^2}=1 \tag{4}$$
Substituting $\alpha=1$ we get:
$$p-q-r=0$$
$\implies$
$$p=0$$
Similarly $$q=r=0$$
But $p=r=0$ is contradicting the result in $(4)$
So is $\alpha=1$ or $\alpha$ is not defined?
 A: The result in (4) is only valid if $p^2+pr+r^2\neq 0$, so that you can divide by it.  So there is no issue with reaching a contradiction from (4); that just is a proof by contradiction that $p^2+pr+r^2=0$.  (Actually, there is no problem reaching a contradiction even without this comment, since that would just prove the hypotheses are impossible and so the conclusion $a=b=c$ is vacuously true.)
To complete your solution, then, you just have to prove that $a=b=c$ if $p^2+pr+r^2=0$.  This is easy: $p^2+pr+r^2=0$ for real $p,r$ implies $p=r=0$ (if $p\neq 0$ you can divide by $p^2$ and find that $r/p$ is nonreal, and similarly if $r\neq 0$).  This then gives $a=b=c$.
A: Your main error occurred while eliminating $\alpha^2$ from equations
$q\alpha^2-r\alpha-p=0$ and $r\alpha^2-p\alpha-q=0$.
Reason:
Once you eliminate $\alpha^2$ we get
$$(r^2-pq)\alpha=p^2-qr$$
Here actually you have taken the term $r^2-pq$ to the denominator without knowing its non zero nature.
Now here is the solution:
In case $r^2-pq=0$ we get $p^2+pr+r^2=0$, But since $p,r$ are reals, this gives $p=0=r$ and so $q=0$ too.
So your equation $(4)$ has no meaning, that means for any $\alpha$ its always true that 
$$(r^2-pq)\alpha=p^2-qr$$
Now in case $r^2-pq \ne 0$, then as you pointed out $\alpha=1$, which gives again $p=q=r=0$ which is contradicting $r^2-pq \ne 0$, hence $r^2-pq$ should be zero.
So finally $r^2-pq=0$ which gives $p=q=r=0$ which implies $a=b=c$ for any $\alpha$.
So essentially value of $\alpha$ is irrelevant here. 
