If $\alpha$ and $\beta$ are the roots of the equation $x^2-px+q=$0,the quadratic equation the roots of which are $\left({\alpha }^2-{\beta }^2\right)\left({\alpha }^3-{\beta }^3\right)$ and ${\alpha }^3{\beta }^2+{\alpha }^2{\beta }^3$

We have,

$\alpha +\beta =p$ and $\alpha \beta =q$

Now, sum of the roots

$\begin{array}{l}=\left({\alpha }^2-{\beta }^2\right)\left({\alpha }^3-{\beta }^3\right)+{\alpha }^3{\beta }^3+{\alpha }^2{\beta }^3\\ =(\alpha -\beta {)}^2(\alpha +\beta )\left({\alpha }^2+\alpha \beta +{\beta }^2\right)+(\alpha \beta {)}^2(\alpha +\alpha )\\ =\left\{(\alpha +\alpha {)}^2-4\alpha \beta \right\}p\left(p^2-q\right)+q^{2}p\\ =\left(p^2-4q\right)p\left(p^2-q\right)+q^{2}p\\ =p\left(\left(p^2-4q\right)\left(p^2-q\right)+q^2\right)\\ =p\left(p^4-5p^{2}q+4q^2\right)\end{array}$

and product of the roots

$=\left({\alpha }^2-{\beta }^2\right)\left({\alpha }^3-{\beta }^3\right)\left(\alpha \beta {)}^2\right(\alpha +\alpha )$

$\begin{array}{l}=\left(p^2-4q^2\right)p\left(p^2-q\right)p{q}^2\\ =p^2{q}^2\left(p^2-4q^2\right)\left(p^2-q\right)\end{array}$

Hence, the required equation is

$x^2-p\left(p^4-5p^{2}q+5q^2\right)x+p^2{q}^2\left(p^2-5p^{2}q+4q^2\right)=0$