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

$q={\alpha }^3+{\beta }^3=(\alpha +\beta {)}^3-3a\beta (\alpha +\beta )$

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

We have

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

$\begin{array}{l}=\frac{(\alpha +\beta {)}^2}{\alpha \beta }-2=\frac{p^2}{\left(p^3+q\right)/3p}-2\\ =\frac{3p^3-2p^3-2q}{p^3+q}=\frac{p^3-2q}{p^3+q}\end{array}$

and $\frac{\alpha }{\beta }\cdot \frac{\beta }{\alpha }=1$

Thus, required quadratic equation is

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