Sum =  $\frac{\mathrm{\alpha }}{\mathrm{\beta }}+\frac{\mathrm{\beta }}{\mathrm{\alpha }}$    or   Sum =  $\frac{{\mathrm{\alpha }}^2+{\mathrm{\beta }}^2}{\mathrm{\alpha \beta }}$ ;  Given  $\mathrm{\alpha }$ + $\mathrm{\beta }$ = – p   and  $\mathrm{\alpha }$³ + $\mathrm{\beta }$³ = q

  ${\mathrm{\alpha }}^2+{\mathrm{\beta }}^2+2\mathrm{\alpha \beta }={\mathrm{p}}^2$                   ....(1)

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

 ${\mathrm{\alpha }}^2+{\mathrm{\beta }}^2-\mathrm{\alpha \beta }=\frac{-\mathrm{q}}{\mathrm{p}}$             .....(2)

(1) - (2) given  $3\mathrm{\alpha \beta }={\mathrm{p}}^2+\frac{\mathrm{q}}{\mathrm{p}}=\frac{{\mathrm{p}}^3+\mathrm{q}}{\mathrm{p}}\Rightarrow \mathrm{\alpha \beta }=\frac{{\mathrm{p}}^3+\mathrm{q}}{3\mathrm{p}}$

required equation is  ${\mathrm{x}}^2-\left(\frac{{\mathrm{\alpha }}^2+{\mathrm{\beta }}^2}{\mathrm{\alpha \beta }}\right)\mathrm{x}+1=0$

from (2), we get   $\frac{{\mathrm{\alpha }}^2+{\mathrm{\beta }}^2}{\mathrm{\alpha \beta }}=\frac{-\mathrm{q}}{\mathrm{p\alpha \beta }}+1=\frac{-3\mathrm{pq}}{\mathrm{p}({\mathrm{p}}^3+\mathrm{q})}+1=\frac{{\mathrm{p}}^3-2\mathrm{q}}{{\mathrm{p}}^3+\mathrm{q}}$

equation  ${\mathrm{x}}^2-\frac{({\mathrm{p}}^3-2\mathrm{q})}{{\mathrm{p}}^3+\mathrm{q}}\mathrm{x}+1=0$

(p³ + q)x² – (p³ – 2q)x + (p³ + q) = 0