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

Given, $$\alpha$$ + $$\beta$$ = $$-$$ p and $$\alpha$$³ + $$\beta$$³ = q

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

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

and $${(\alpha + \beta )^2} = {p^2}$$

$$ \Rightarrow {\alpha ^2} + {\beta ^2} + 2\alpha \beta = {p^2}$$ ..... (ii)

From Eq. (i) and (ii), we get

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

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

$$\therefore$$ Required equation is

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

$$ \Rightarrow ({p^3} + q){x^2} - ({p^3} - 2q)x + ({p^3} + q) = 0$$