Sum of roots $=\frac{\alpha^2+\beta^2}{\alpha \beta}$ and product $=1$
Given, $\alpha+\beta=-p$ and $\alpha^3+\beta^3=q$
$$
\begin{aligned}
& \Rightarrow(\alpha+\beta)\left(\alpha^2-\alpha \beta+\beta^2\right)=q \\
& \therefore \quad \alpha^2+\beta^2-\alpha \beta=\frac{-q}{p}
\end{aligned}
$$
and $\quad(\alpha+\beta)^2=p^2$
$$
\Rightarrow \quad \alpha^2+\beta^2+2 \alpha \beta=p^2
$$
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
$$
\begin{gathered}
x^2-\frac{\left(p^3-2 q\right) x}{\left(p^3+q\right)}+1=0 \\
\Rightarrow\left(p^3+q\right) x^2-\left(p^3-2 q\right) x+\left(p^3+q\right)=0
\end{gathered}
$$