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

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.
