If $x^2+px+1$ is a factor of $ax^3+bx+c$ then which of the following is true:
a)$a^2-b^2=ac$
b)$a^2-c^2=ab$
c)$a^2+b^2=bc$
d)$a^2+c^2=ab$
I assumed the roots of $x^2+px+1$ to be $k$ and $\frac{1}{k}$ since product of roots is 1 . $ax^3+bx+c$ would have roots $k$, $\frac{1}{k}$ and $r$. So $$(k)( \frac{1}{k})(r)=-c \Rightarrow -c=r$$ So $-c$ is a root and hence $$a(-c)^3+b(-c)+c=0 \Rightarrow -ac^2-b+1=0$$ as $c\neq0$ but there is no such solution so how should I proceed??
$ax^3+bx+c$ is the product of $x^2+px+1$ and a linear polynomial. This polynomial must be $ax+c$.
$$ax^3+bx+c=(ax+c)(x^2+px+1)$$
So, $c+ap=0$ and $a+pc=b$.
$$0=c(c+ap)=c^2+acp=c^2+a(b-a)$$