Four roots of the polynomial $x^4+px^3​+qx^2​+rx+1$ Let $a, b, c, d$ be four roots of the polynomial $x^4+px^3​+qx^2​+rx+1$ prove that $(a^4​+1)(b^4​+1)(c^4​+1)(d^4​+1)=(p^2​+r^2)​^2​+q^4-4pq^2​r$
Please provide hint. 

Let


*

*$f(x) = x^4 + px^3 + qx^2 + rx + 1$.

*$\Lambda = \{ a, b, c, d \}$ be the set of roots of $f(x)$.

*$\Omega = \left\{ \frac{\pm 1 \pm i}{\sqrt{2}} \right\}$ be the set of roots of $x^4 + 1$. 


We have
$$\begin{align}\mathcal{P}\stackrel{def}{=}\prod_{\lambda \in \Lambda}(\lambda^4 + 1)
&= \prod_{\lambda \in \Lambda}\prod_{\omega \in \Omega}(\lambda - \omega)
= \prod_{\omega \in \Omega}\prod_{\lambda _in \Lambda}(\omega - \lambda)
= \prod_{\omega \in \Omega}f(\omega)\\
&= \prod_{\omega \in \Omega}(p \omega^3 + q \omega^2 + r\omega)
\end{align}
$$
Since $\prod_{\omega \in \Omega} \omega^2 = (-1)^2 = 1$ and $\omega^{-1} = \bar{\omega}$ for all $\omega \in \Omega$, we find
$$\begin{align}
\mathcal{P} = & \prod_{\omega \in \Omega}(q + p \omega + r \bar\omega)\\
= & \phantom{+}
\left(q + \frac1{\sqrt{2}}(p + r) + \frac{i}{\sqrt{2}}(p - r)\right)
\left(q + \frac1{\sqrt{2}}(p + r) - \frac{i}{\sqrt{2}}(p - r)\right)\\
& \times 
\left(q - \frac1{\sqrt{2}}(p + r) + \frac{i}{\sqrt{2}}(p - r)\right)
\left(q - \frac1{\sqrt{2}}(p + r) - \frac{i}{\sqrt{2}}(p - r)\right)\\
= & \left(q^2 + p^2 + r^2 + \sqrt{2}q(p+r)\right)
\left(q^2 + p^2 + r^2 - \sqrt{2}q(p+r)\right)\\
= & (q^2 + p^2 + r^2)^2 - 2q^2(p+r)^2\\
= & (p^2 + \color{red}{r}^2)^2 + q^4 - 4q^2 pr
\end{align}$$
Please note that there is a typo in the formula you want to prove.
