If $\alpha, \beta, \gamma, \delta$ be the roots of the equation $x^4+px^3+qx^2+rx+s=0$, Show that the equation whose roots are $(\alpha \beta + \gamma \delta), (\beta \gamma + \alpha \delta), (\gamma \alpha + \beta \delta)$ is $$x^3-qx^2+(pr-4s)x-(r^2-4qs+p^2s)=0$$
I'm really stuck on this one. I've tried using simple manipulations, methods for finding equations with symmetric functions of roots, Newton's theorem. But, nothing seems to work. Any kind of help is appreciated.
$\alpha,\beta,\gamma,\delta$ being the roots of the quartic, they verify the following
$$\begin{align} \alpha+\beta+\gamma+\delta&=-p\\ \alpha\beta+\alpha\gamma+\alpha\delta+\beta\gamma+\beta\delta+\gamma\delta&=q\\ \alpha\beta\gamma+\alpha\beta\delta+\alpha\gamma\delta+\beta\gamma\delta&=-r\\ \alpha\beta\gamma\delta&=s \end{align}$$
Now denote $A=\alpha\beta+\gamma\delta$, $B=\beta\gamma+\alpha\delta$ and $\Gamma=\gamma\alpha+\beta\delta$. These three numbers are the roots of the following cubic.
$$X^3-(A+B+\Gamma)X^2+(AB+A\Gamma+B\Gamma)X-AB\Gamma=0$$
A straightforward computation gives
$$\begin{align} A+B+\Gamma&=\alpha\beta+\alpha\gamma+\alpha\delta+\beta\gamma+\beta\delta+\gamma\delta=q\\ AB+A\Gamma+B\Gamma&=pr-4s\\ AB\Gamma&=r^2-4qs+p^2s \end{align}$$