If $x^2 +px +1$ is a factor of $ ax^3 +bx+c$ then relate $a,b,c$ Suppose If $x^2 +px +1$ is a factor of $ax^3 +bx+c$ then relate $a,b,c$ such that $a,b,c \in R$
I can write $$ax^3 +bx+c=(x^2 +px +1)(\lambda x +D)$$
$$\implies ax^3 +bx+c =\lambda x^3 + x^2.p\lambda + x(\lambda+pD)+D $$
and then compare coefficient to find out relation but that will be long and tedious process , I want shorter approach to this problem . Btw I was given following options for this question

A) $a^2+c^2+ab=0$
B)  $a^2-c^2+ab=0$
C) $a^2-c^2-ab=0$
D) $ap^2+bp+c=0$

Maybe we can relate something by looking at options?

Following the path you have taken, and correcting the algebra, one Gets
$$(\lambda X+D)(X^2+pX+1)=\lambda X^3+(\lambda p+D)X^2+(Dp+\lambda)X+D$$
And this is valid for all $X$. So by identifying the coefficients we get
$$\begin{cases} \lambda=a\\ \lambda p+D=0\\ Dp+\lambda=b\\ D=c \end{cases}$$
which reads
$$\begin{cases} a p+c=0\\ cp+a=b\end{cases}$$
and this leads to
$$-\frac{c^2}{a}+a=b$$
$$a^2-c^2-ab=0$$
