Vieta's formulas for $x^2+px+1$ and $x^2+qx+1$. 
Let $\alpha$ and $\beta$ be the roots for $x^2+px+1$ and let $\gamma$ and $\delta$ be the roots
  for $x^2+qx+1$. Show that $$(\alpha-\gamma)(\beta-\gamma)(\alpha+\delta)(\beta+\delta)=q^2-p^2.$$

This seemed to be rather peculiar. It should be a simple application of Vieta's formulas, but I couldn't get it to the form they wanted...
We have $\alpha+\beta=-p$, $\alpha\beta=1$ and $\gamma+\delta=-q$, $\gamma\delta=1$.
Also $\alpha^2+p\alpha+1=0 \Rightarrow p = -\alpha-\frac{1}{\alpha} \Rightarrow p^2 = \alpha^2+2+ \frac{1}{\alpha^2}$
and $\gamma^2+q\gamma+1=0 \Rightarrow p = -\gamma-\frac{1}{\gamma} \Rightarrow q^2 = \gamma^2+2+ \frac{1}{\gamma^2}$.
This should imply that $q^2-p^2 = (\gamma^2+2+\frac{1}{\gamma^2})-(\alpha^2+2+ \frac{1}{\alpha^2}) = \gamma^2+\frac{1}{\gamma^2} - \alpha^2-\frac{1}{\alpha^2}$.
What should I do here?

Note $x^2+px+1= (x-\alpha)(x-\beta)$. Then
\begin{align}
& (\alpha-\gamma)(\beta-\gamma)(\alpha+\delta)(\beta+\delta)\\
= &(\gamma^2+p\gamma +1)(\delta^2-p\delta +1)\\
= &(\gamma\delta)^2+\delta^2 +\gamma^2+1-(\gamma\delta)p^2\\
 = &(\delta+\gamma)^2-2\gamma\delta+2-p^2\\
=&q^2-p^2
\end{align}
where $\gamma+\delta=-q $ and $\gamma\delta=1$.
