Common linear and Quadratic factors (i) If $ax^5+bx^2+c$ has a factor of the form $x^2+px+1$ prove that:

$(a^2-c^2)(a^2-c^2+bc)=(ab)^{2}$

(ii)In this case prove that:
$ax^5+bx^2+c$ and $cx^5+bx^3+a$ have a common quadratic factor

This is a sequel to my previous work. The aim is to get the relation $(a^2-c^2)(a^2-c^2+bc)=(ab)^{2}$ directly rather than just verifying it. In order to do so, I have to re-arrange the original question as
“If $ax^5+bx^2+c$ and $cx^5+bx^3+a$ have a common quadratic factor $x^2+px+1$, then $(a^2-c^2)(a^2-c^2+bc)=(ab)^{2}$.”
[I can do that because the second part of the original question has been proved independently.]
For some polynomials P(x) and Q(x) of degree 3, we let
$f(x) = ax^5 + bx^2 + c = (x^2 + px + 1)P(x)…………..(1)$
$g(x) = cx^5 + bx^3 + a = (x^2 + px + 1)Q(x)…………..(2)$
Let $m$ and $n$ (where $m.n = 1$) be the roots of $x^2+ px +1= 0$, then m is also a root of $f(x) = 0$ and of $g(x) = 0$. Thus,
$ am^5 + bm^2 + c = 0…..........(1.1)$
$ cm^5 + bm^3 + a = 0…......….(2.1)$
Do [a*(2.1) – c*(1.1)] and get $abm^3 – cbm^2 + (a^2 – c^2) = 0$
This means m is also a root of $F(x) = abx^3 – cbx^2 + (a^2 – c^2) = 0$ if we define
$F(x) = a*g(x) – c*f(x)$
On one hand $F(x) = abx^3 – cbx^2 + (a^2 – c^2)$
On the other hand $F(x) = (x^2+px+1).[aQ(x)] – (x^2+px+1)[cP(x)] = (x^2+ px+1).[R(x)$, a polynomial of degree 1 in x]
i.e. $F(x) = abx^3 – cbx^2 + (a^2 – c^2) = (x^2 + px + 1).[R(x)]$
Using the similar reasoning as before, $(a^2 – c^2)x^3 – cbx^1 + ab$ also has the same quadratic factor $(x^2+ px+1)$
For some polynomial, S(x) of degree 1 in x, define $G(x) = (a^2 – c^2)x^3 – cbx^1 + ab = (x^2 + px + 1).[S(x)]$
Since m is a root of $x^2 + px + 1 = 0$, m is also a root of $G(x) = (a^2 – c^2)x^3 – cbx^1 + ab = 0$
From $(a^2 – c^2)*F(x) – (ab)*G(x) = 0$, we have  
$$x^2 – \frac {ab^2c} {bc(a^2 – c^2)}x + \frac {a^2b^2 – (a^2 – c^2)} {bc(a^2 – c^2)} = 0$$
The LHS of the above is actually that quadratic factor.
Hence, $[a^2b^2 – (a^2 – c^2)] / [bc(a^2 – c^2)] = 1$
The required relation in a, b and c then follows.
