Let $p$ and $q$ be real numbers such that $p\ne 0,p^3\ne -q.$ If $\alpha$ and $\beta$ are non-zero complex numbers satisfying $\alpha +\beta =-p$ and ${\alpha }^3+{\beta }^3=q\text{, }$a quadratic equation having $\frac{\alpha }{\beta }$ and $\frac{\beta }{\alpha }$ as its roots, is 

$\frac{\alpha }{\beta }+\frac{\beta }{\alpha }=\frac{{\alpha }^2+{\beta }^2}{\alpha \beta }=\frac{(\alpha +\beta {)}^2-2\alpha \beta }{\alpha \beta }$               …(i)

and given           ${\alpha }^3+{\beta }^3=q,\alpha +\beta =-p$

$\begin{array}{l}\Rightarrow (\alpha +\beta {)}^3-3\alpha \beta (\alpha +\beta )=q\\ \Rightarrow -p^3+3p\alpha \beta =q\end{array}$

or                                $\alpha \beta =\frac{q+p^3}{3p}$

$\therefore$ From Eq. (i), we get

           $\frac{\alpha }{\beta }+\frac{\beta }{\alpha }=\frac{p^2-\frac{2(q+p^3)}{3p}}{\frac{\left(q+p^3\right)}{3p}}=\frac{p^3-2q}{\left(q+p^3\right)}$

and product of the roots $=\frac{\alpha }{\beta }\cdot \frac{\beta }{\alpha }=1$

$\therefore$Required equation is $x^2-\left(\frac{p^3-2q}{q+p^3}\right)x+1=0$

 or   $\left(q+p^3\right)x^2-\left(p^3-2q\right)x+\left(q+p^3\right)=0$