jmc

Since $\alpha$ and $\beta$ are the roots of $x^2+px+1=0,$ $$(x-\alpha )(x-\beta )=x^2+px+1.$$Setting $x=\gamma ,$ we get $$(\gamma -\alpha )(\gamma -\beta )={\gamma }^2+p\gamma +1.$$or $(\alpha -\gamma )(\beta -\gamma )={\gamma }^2+p\gamma +1.$

Setting $x=-\delta ,$ we get $$(-\delta -\alpha )(-\delta -\beta )={\delta }^2-p\delta +1,$$or $(\alpha +\beta )(\beta +\delta )={\delta }^2-p\delta +1.$

Since $\gamma$ and $\delta$ are the roots of $x^2+qx+1=0,$ ${\gamma }^2+q\gamma +1=0$ and ${\delta }^2+q\delta +1=0.$ Then $${\gamma }^2+p\gamma +1=(p-q)\gamma$$and $${\delta }^2-p\delta +1=-(p+q)\delta .$$Finally, by Vieta's formulas, $\gamma \delta =1,$ so $$(p-q)\gamma \cdot (-(p+q))\delta =(q-p)(q+p)=\boxed{q^2-p^2}.$$