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
 A: Proving part (ii) independently is less tedious than part (i). Here it is:-
Let $f(x) = ax^5 + bx^2 + c$ 
Also $f(x) = (x^2 + px + 1)Q_3 (x)$; where $Q_3 (x)$ is a quotient polynomial of degree 3…….(1)
Let $g(x) = x^2 + px + 1$ such that g(x) = 0 has m and n as roots.
From product of roots, $n = 1/m$.
$f(n) = f(1/m) = … = (1/m)^5 . [a + b(m)^3 + c(m)^5]…....(2)$
From (1), $f(n) = f(1/m) = … = (1/m)^2 . [1 + p(m) + (m)^2]Q_3 (1/m)……(3)$
Equating (2) and (3), we have $a + bm^3 + cm^5 = (1 + pm + m^2).[m^3.Q_3 (1/m)]$
This means $a + bm^3 + cm^5$ has $(1 + pm + m^2)$ as factor.
i.e. $cx^5 + bx^3 + a$ has $(x^2 + px +1)$ as factor.
Thus, $ax^5+bx^2+c$ and $cx^5+bx^3+a$ have a common quadratic factor.
For part(i), I still need some time to workout.
A: Part (i) is a continuation of Part (ii).
m is a root of g(x) = 0 and hence is a root of f(x) = 0.
Therefore, $am^5 + bm^2 + c = 0$……(4)
Using n = 1/m as root, from part (ii), we have $a + bm^3 +cm^5 = 0$…….(5)
Eliminating a from (4) & (5) by performing $(5)*m^5 – (4)$, we have
$bm^2(m^6 – 1) + c[m^T – 1] = 0$……(6) [T = 10 because of the display matter.]
Using the same technique to eliminate b and then c, we get
$a(m^6 – 1) – cm(m^4 – 1) = 0$……..(7)
$a(m^T – 1) + bm^3(m^4 – 1) = 0$ …...(8)
(6), (7), (8) combined yield the ratio of a : b : c such that, for some k
$a = –m^3(m^4 – 1) k$;
$b = (m^T – 1)k$; and
$c = –m^2(m^6 – 1)k$.
Using these values can verify that LHS = … = $m^6k^4(m^T – 1)^2(1 – m^4)^2$ = … = RHS.
Note-1 Who designs this question that involves so many calculations?
Note-2 This work is just doing the verification. How to get the required product is still unknown.
Note-3 Would like to know any other shorter method or better approach.
A: 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.
