Given that \( \Large x^{2}+px+1 \) is factor of \( \Large ax^{3}+bx+c=0 \) then

let \( \Large ax^{3}+bx+c= \left(x^{2}+px+1\right) \left(ax+\lambda\right) \), where \( \Large \lambda \) is constant. Then equating the coefficient of like powers of x on both sides, we get.

\( \Large 0 = ap+\lambda, b=p\lambda+a, c=\lambda \)

\( \Large p=\frac{-\lambda}{a}=\frac{-c}{a} \)

Hence, \( \Large b= \left(-\frac{c}{a}\right)c+a \) or \( \Large ab=a^{2}-c^{2} \)