Let p and q be real numbers such that $p\ne 0$ $p^3\ne q\text{ and }{p}^3\ne -q\text{. }$If $a$ and $\beta$ are non-zero complex numbers satisfying $\alpha +\beta =-\rho$ and 

${\alpha }^3+{\beta }^3=q$ then a quadratic equation having $\frac{\alpha }{\beta }\text{ and }\frac{\beta }{\alpha }$ as its roots is

Sum of roots $=\frac{{\alpha }^2+{\beta }^2}{\alpha \beta }$ and product of roots= 1 

Given,  $\alpha +\beta =-\rho$

and ${\alpha }^3+{\beta }^3=0$

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

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

and $(\alpha +\beta {)}^2={\rho }^2$

$\Rightarrow {\alpha }^2+{\beta }^2+2\alpha \beta ={\rho }^2$   …(ii)

From Eqs. (i) and (ii), we get 

${\alpha }^2+{\beta }^2=\frac{p^3-2q}{3p}$

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

$\therefore$ Required equation is 

$\begin{array}{l}{x}^2-\frac{\left(p^3-2q\right)x}{\left(p^3+q\right)}+1=0\\ \Rightarrow \left(p^3+q\right)x^2-\left(p^3-2q\right)x+\left(p^3+q\right)=0\end{array}$