If $\alpha$ and $\beta$ be the roots of ${\mathrm{x}}^2+\mathrm{px}+\mathrm{q}=0$ then, $\frac{\left(\mathrm{\omega \alpha }+{\mathrm{\omega }}^2\mathrm{\beta }\right)\left({\mathrm{\omega }}^2\mathrm{\alpha }+\mathrm{\omega \beta }\right)}{\frac{{\mathrm{\alpha }}^2}{\mathrm{\beta }}+\frac{{\mathrm{\beta }}^2}{\mathrm{\alpha }}} \mathrm{isequal} \mathrm{to}$

Since, $\alpha$ and $\beta$ are the roots of the equation ${\mathrm{x}}^2+\mathrm{px}+\mathrm{q}=0$, therefore$\mathrm{\alpha }+\mathrm{\beta }=-\mathrm{p}\text{ and }\mathrm{\alpha \beta }=\mathrm{q}$, Now, $\left(\mathrm{\omega \alpha }+{\mathrm{\omega }}^2\mathrm{\beta }\right)\left({\mathrm{\omega }}^2\mathrm{\alpha }+\mathrm{\omega \beta }\right)$

$\begin{array}{l}={\mathrm{\alpha }}^2+{\mathrm{\beta }}^2+\left({\mathrm{\omega }}^4+{\mathrm{\omega }}^2\right)\mathrm{\alpha \beta }\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\left(\because {\mathrm{\omega }}^3=1\right)\\ ={\mathrm{\alpha }}^2+{\mathrm{\beta }}^2-\mathrm{\alpha \beta }\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\left(\because \mathrm{\omega }+{\mathrm{\omega }}^2=-1\right)\\ =(\mathrm{\alpha }+\mathrm{\beta }{)}^2-3\mathrm{\alpha \beta }={\mathrm{p}}^2-3\mathrm{q}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\end{array}$

Also, $\begin{array}{r}\frac{{\mathrm{\alpha }}^2}{\mathrm{\beta }}+\frac{{\mathrm{\beta }}^2}{\mathrm{\alpha }}=\frac{{\mathrm{\alpha }}^3+{\mathrm{\beta }}^3}{\mathrm{\alpha \beta }}\\ =\frac{(\mathrm{\alpha }+\mathrm{\beta }{)}^3-3\mathrm{\alpha \beta }(\mathrm{\alpha }+\mathrm{\beta })}{\mathrm{\alpha \beta }}=\frac{\mathrm{p}\left(3\mathrm{q}-{\mathrm{p}}^2\right)}{\mathrm{q}}\end{array}$

$\therefore \text{ The given expression }=\frac{\left({\mathrm{p}}^2-3\mathrm{q}\right)}{\frac{\mathrm{p}\left(3\mathrm{q}-{\mathrm{p}}^2\right)}{\mathrm{q}}}=-\frac{\mathrm{q}}{\mathrm{p}}$