A problem involving the roots of quartic polynomial $x^4+px^3+qx^2+rx+1$

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,

$$(1+\alpha_1^4)(1+\alpha_2^4)(1+\alpha_3^4)(1+\alpha_4^4) =f(-1)=(p^2+r^2)^2+q^4-4pq^2r $$

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

$\alpha^4+1=(\eta-\alpha)(\eta^2-\alpha)(\eta^5-\alpha)(\eta^7-\alpha)$ 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.