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?
 A: 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 $\gamma\delta=1$.
A: You can solve it like this e.g.   
Sorry for the image but it would take me 1 hour to type this in MathJax.   

A: Regrouping the factors, one has $$(\beta-\gamma)(\alpha+\delta)\cdot(\alpha-\gamma)(\beta+\delta)$$
$$=(\beta\delta-\gamma\alpha)(\alpha\delta-\gamma\beta)$$
$$=\delta^2-\beta^2-\alpha^2+\gamma^2$$
$$=(\delta+\gamma)^2-2\delta\gamma-(\beta+\alpha)^2+2\beta\alpha$$
$$=(\delta+\gamma)^2-(\beta+\alpha)^2$$
$$=q^2-p^2,$$ where one uses $\alpha\beta=\gamma\delta=1$ during the process.
Note. Your last step equals $$\gamma^2+\delta^2-\alpha^2-\beta^2$$ which can be simplified as above.
A: As $\alpha\beta=\gamma\delta$
assuming  $\alpha\beta\gamma\delta$ are non-zero & finite,
let $\dfrac\alpha\gamma=\dfrac\delta\beta=k$(say)
$$-p=\alpha+\beta\implies \beta+k\gamma+p=0\  \ \ \  (1)$$
$$\text{Similarly, }\beta k+\gamma+q=0\  \ \ \  (2)$$
Find $(1)+(2), (1)-(2)$
Hope you can take it from here?
