I've seen it before - 4

If $x^2 + px +1$ is factor of $ax^3 + bx +c$ , then:

$ax^3 + bx +c = (ax-q)(x^2 + px +1)$

$\quad \quad \quad = ax^3 + (ap-q)x^2 + (a-pq)x - q$

$\Rightarrow ap-q = 0, b = a - pq, c = -q$

$\Rightarrow ap = q = -c \quad \Rightarrow p = - \dfrac {c} {a}$

$\Rightarrow b = a - pq = a - \left( \dfrac {-c}{a} \right) (-c) = a - \dfrac {c^2} {a}$

$\Rightarrow a - \dfrac {c^2} {a} = b \quad \Rightarrow \boxed {a^2 - c^2 = ab}$

If $x_1,x_2,x_3$ be the roots, we can consider any two among them, say $x_1,x_2$ to be the roots of the quadratic factor.

so, $x_1+x_2+x_3=0$ and $x_1+x_2=-p$ ,

and, $x_1x_2=1$ and $x_1x_2x_3=-\frac{c}{a}$

solving, we get: $x_3=p=-\frac{c}{a}$

So, we can write the cubic as

$a(x+\frac{c}{a})(x^{2}-\frac{c}{a}x+1)=0$

Simplifying, we get:

$ax^{3}+x\frac{(a^{2}-c^{2})}{a}+c=0$

comparing with the original cubic,

$\boxed{a^{2}-c^{2}=ab}$

---

This is a simple, yet nice problem :D

I just paired 1 and C, and thought out the positive equatorial of both. Then I realized the first also had to be positive. A positive plus a positive is a positive, so I chose that one.

since x^2 +px +1 is a factor therefore after long division the reminder should be zero the reminder= (b-a+ap^2)x +(c+ap) =0 this gives u 2 equations: b-a+ap^2 =0 & c+ap=0 solving them together eliminating "p" gives u the answer a^2-c^2=ab