If x2+px+1 is a factor of the expression ax3+bx+c, then
a2−c2=ab
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 costant. Then equating the coefficient of like powers of x on both sides, we get
0=ap+λ,b=pλ+a,c=λ
⇒ p=−λa=−ca
Hence b = (−ca)c+a
⇒ab=a2−c2