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)&amp;=(\alpha-1)(\beta-1)(\gamma-1)(\delta-1)=1+p+q+r+s\\P(-1)&amp;=(\alpha+1)(\beta+1)(\gamma+1)(\delta+1)=1-p+q-r+s.\end{align}

Then \begin{align}P(1)P(-1)&amp;=(\alpha^2-1)(\beta^2-1)(\gamma^2-1)(\delta^2-1)\\&amp;=(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.
You correctly computed 
$$
(\alpha^2-1)(\beta^2-1)(\gamma^2-1)(\delta^2-1)=(1+q+s)^2-(p+r)^2
$$
only that this is not what was asked for.

But you were close: $\alpha^2+1 = (\alpha-i)(\alpha+i)$ etc, and therefore
$$
 (1+\alpha^2)(1+\beta^2)(1+\gamma^2)(1+\delta^2) = P(i) P(-i)
$$
which expands to
$$
(1-ip-q+ir+s)(1+ip-q-ir+s) = (1-q+s)^2+(p-r)^2 \, .
$$
Your approach ends up proving $(α^2 - 1)(β^2 - 1)(γ^2 - 1)(δ^2 - 1) = (1 - q + s)^2 - (p - r)^2$. This means we will have to do something with that minus sign and to do that we can use the identity $(x + ιy)(x - ιy) = x^2 + y^2$. Putting $P(ι)$, and $P(-ι)$ you can easily solve the problem with the same approach.