Explanation

(c)
Given that $x^{2} + px + 1$ is a factor of $ax^{3} + bx + c = 0$, then let $ax^{3} + bx + c \equiv \left( x^{2} + px + 1 \right)(ax + \lambda),\ where\ \lambda$ is a constant. Then, equating the coefficients of like powers of $x$ on both sides, we get
$0 = ap + \lambda,\ b = p\lambda + a,c = \lambda$
$\Rightarrow p = - \frac{\lambda}{a} = - \frac{c}{a}$
Hence, $b = \left( \ - \frac{c}{a} \right)c + a\ \Rightarrow ab = a^{2} - c^{2}$