A problem involving the roots of quartic polynomial $x^4+px^3+qx^2+rx+1$ Let $\alpha_1$, $\alpha_2$, $\alpha_3$, $\alpha_4$ be the roots of the following polynomial
$$P(x)=x^4+px^3+qx^2+rx+1$$
Show that
$$(1+{\alpha_1}^4)(1+{\alpha_2}^4)(1+{\alpha_3}^4)(1+{\alpha_4}^4)=(p^2+r^2)^2+q^4-4pq^2r.$$

I came across this problem on a facebook page as an challenge.The only way that Strike me is multiply the terms in LHS and put the corresponding values but this method would be very long as polynomial is of 4th degree.
If anybody know any other method then Please tell me. Thank you for your help!
 A: Rewrite $x^4+px^3+qx^2+rx+1=0$ as
$$x^2+\frac1{x^2}+q=-(px +\frac r x) $$
Square to get
$$x^4+\frac1{x^4} + (2+q^2-2pr)=(p^2-2q)x^2+ \frac{r^2-2q}{x^2}$$
Square again and rearrange to obtain the quartic equation in $x^4$
\begin{align}
f(x^4)=&x^{16}+[2(2+q^2-2pr)-(p^2-2q)^2]x^{12}\\
&+[2+(2+q^2-2pr)^2-2(p^2-2q)(r^2-2q)]x^8\\
&+[2(2+q^2-2pr)-(r^2-2q)^2]x^4+1=0
\end{align}
Then,
$$(1+\alpha_1^4)(1+\alpha_2^4)(1+\alpha_3^4)(1+\alpha_4^4)
=f(-1)=(p^2+r^2)^2+q^4-4pq^2r
$$
A: Squaring both sides
$$(x^4+qx^2+1)^2=x^2(px^2+r)^2$$
Let $x^2=y$
$$(y^2+qy+1)^2=y(py+r)^2$$
$$\iff y^4+y^2A+1=y^3B+yC$$
Let $z=1+y^2\implies y=\sqrt{z-1}$
$$\implies(z-1)^2+(z-1)A+1=\pm\sqrt{z-1}((z-1)B+C)$$
$$ z^2+zE+F=\pm\sqrt{z-1}(zG+H)$$
Squaring both sides
$$z^4+\cdots+F^2=(z-1)(\cdots+H^2)$$
$$z^4+\cdots+F^2+H^2=0$$
$$\implies\prod_{r=1}(1+\alpha_r^4)=\prod_{r=1}z_r=\dfrac{F^2+H^2}1$$  using Vieta's formula
A: $\alpha^4+1=(\eta-\alpha)(\eta^2-\alpha)(\eta^5-\alpha)(\eta^7-\alpha)$
where $\eta=\exp(\pi i/4)=\frac1{\sqrt2}(1+i)$. Therefore
$$\prod_{k=1}^4(1+\alpha_k^4)=P(\eta)P(\eta^3)P(\eta^5)P(\eta^7).$$
But
$$P(\eta)P(\eta^5)=P(\eta)P(-\eta)$$
and
$$P(x)P(-x)=(x^4+qx^2+1)^2-(px^3+rx)^2
=Q(x^2)$$
where
$$Q(x)=x^4+(2q-p^2)x^3+(2+q^2-2pr)x^2+(2q-r^2)x+1.$$
So
$$P(\eta)P(\eta^5)=Q(\eta^2)=Q(i)$$
and similarly
$$P(\eta^3)P(\eta^7)=Q(-i).$$
Then
$$\prod_{k=1}^4(1+\alpha_k^4)=Q(i)Q(-i)=(2pr-q^2)^2+(p^2-r^2)^2.$$
This should be re-arrangeable into your answer.
