If x2+px+1 is a factor of the expression ax3+bx+c, then

Given that x2+px+1 is factor of ax3+bx+c=0, then let ax3+bx+c=(x2+px+1)(ax+λ), where λ is a constant. Then equating the coefficients of the powers of x on both sides, we get0=ap+λ, b=pλ+a, c=λ⇒p=-λa=-caHence b=(-ca)c+a or ab=a2-c2 .