If $p$ and $q$ are non-zero real numbers and $\alpha^3+\beta^3=-p, \alpha \beta=q$, then a quadratic…
If $p$ and $q$ are non-zero real numbers and $\alpha^3+\beta^3=-p, \alpha \beta=q$, then a quadratic equation whose roots are $\frac{\alpha^2}{\beta}, \frac{\beta^2}{\alpha}$ is :
$p x^2-q x+p^2=0$
$q x^2+p x+q^2=0$
$p x^2+q x+p^2=0$
$q x^2-p x+q^2=0$
Solution
Given $\alpha^3+\beta^3=-p$ and $\alpha \beta=q$
Let $\frac{\alpha^2}{\beta}$ and $\frac{\beta^2}{\alpha}$ be the root of required quadratic equation.
So, $\frac{\alpha^2}{\beta}+\frac{\beta^2}{\alpha}=\frac{\alpha^3+\beta^3}{\alpha \beta}=\frac{-p}{q}$
and $\frac{\alpha^2}{\beta} \times \frac{\beta^2}{\alpha}=\alpha \beta=q$
Hence, required quadratic equation is
$
\begin{aligned}
&x^2-\left(\frac{-p}{q}\right) x+q=0 \\
&\Rightarrow x^2+\frac{p}{q} x+q=0 \Rightarrow q x^2+p x+q^2=0
\end{aligned}
$