If $\alpha$, $\beta$, $\gamma$ and $\delta$ be the roots of the polynomial $x^4+px^3+qx^2+rx+s$, prove the following. 
If $\alpha$, $\beta$, $\gamma$ and $\delta$ be the roots of the polynomial $x^4+px^3+qx^2+rx+s$, show that $$(1+\alpha^2)(1+\beta^2)(1+\gamma^2)(1+\delta^2)=(1-q+s)^2+(p-r)^2.$$

I have tried putting
\begin{align}P(1)&=(\alpha-1)(\beta-1)(\gamma-1)(\delta-1)=1+p+q+r+s\\P(-1)&=(\alpha+1)(\beta+1)(\gamma+1)(\delta+1)=1-p+q-r+s.\end{align}
Then \begin{align}P(1)P(-1)&=(\alpha^2-1)(\beta^2-1)(\gamma^2-1)(\delta^2-1)\\&=(1+p+q+r+s)(1-p+q-r+s)=(1+q+s)^2-(p+r)^2\end{align}
Somehow it does not match the statement given.

Another way using Transformation of equation: 
We have $$x^4+qx^2+s=-x(px^2+r)$$
Squaring both sides
$$(x^2)^4+q^2(x^2)+s^2+2q(x^2)^3+2qs(x^2)+2s(x^2)^2=x^2(p^2(x^2)^2+r^2+2pr(x^2))$$
Replacing $x^2+1=y$
$$(y-1)^4+q^2(y-1)^2+s^2+2q(y-1)^3+2qs(y-1)+2s(y-1)^2=(y-1)(p^2(y-1)^2+r^2+2pr(y-1))$$
$$\iff y^4+\cdots+1+q^2+s^2-2q-2qs+2s+p^2+r^2-2pr=0$$
Using Vieta's formula $$\prod_{j=1}^4y_j=\dfrac{1+q^2+s^2-2q-2qs+2s+p^2+r^2-2pr}1=(1-q+s)^2+(p-r)^2$$
