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

Sum of roots \(=\frac{\alpha^2+\beta^2}{\alpha \beta}\) and product \(=1\)
Given, \(\alpha+\beta=-p\) and \(\alpha^3+\beta^3=q\)
\(\Rightarrow(\alpha+\beta)\left(\alpha^2-\alpha \beta+\beta^2\right)=q \)
\( \therefore \alpha^2+\beta^2-\alpha \beta=\frac{-q}{p}\)
and \(\quad(\alpha+\beta)^2=p^2\)
\(
\Rightarrow \alpha^2+\beta^2+2 \alpha \beta=p^2
\)
From Eqs. (i) and (ii), we get
\(
\alpha^2+\beta^2=\frac{p^3-2 q}{3 p}
\)
and \(\alpha \beta=\frac{p^3+q}{3 p}\)
\(\therefore\) Required equation is
\( x^2-\frac{\left(p^3-2 q\right) x}{\left(p^3+q\right)}+1=0 \)
\( \Rightarrow\left(p^3+q\right) x^2-\left(p^3-2 q\right) x~+\) \(\left(p^3+q\right)=0 \)