Please answer in detail…………………………………………………....……………

Since x^2+px+1 is a factor of ax^3+bx+c, the other factor should be ax+c. ( by comparing x^3 term and the constant term)

So ax^3+bx+c = (x^2+px+1)(ax+c)

Comapring the x terms and x^2 terms we get

b = pc + a => 2ab = 2pca + 2a^2 …(1)

0 = c + pa. Squaring both sides we get

0 = c^2 + 2pca +a^2 ……(2)

(1) - (2) gives

2ab = 2pca + 2a^2 - c^2 - 2pca - a^2

2ab = a^2 - c^2.