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. 

I use the following trick.

If $f(x)$ is a polynomial, then $f(x)f(-x)$ is a polynomial with only even degree terms, so there exists a polynomial $g(x)$ such that 
  $$g(x^2)=f(x)f(-x).$$ Then the zeros of $g(x)$ are exactly the squares of the zeros of $f(x)$.

With $f(x)=x^4+px^3+qx^2+rx+1$ we arrive at
$$
g(x)=1 + (2 q - r^2) x + (2 + q^2 - 2 p r) x^2 + (-p^2 + 2 q) x^3 + x^4.
$$
Repeating the dose we look for a polynomial $h(x)$ such that $h(x^2)=g(x)g(-x)$.
Expanding gives us
$$
\begin{aligned}
h(x)&=1 + (4 - 2 q^2 - 4 p r + 4 q r^2 - r^4) x\\
& + (6 + 4 p^2 q - 4 q^2 + 
    q^4 - 8 p r - 4 p q^2 r + 2 p^2 r^2 + 4 q r^2) x^2\\
& + (4 - p^4 + 
    4 p^2 q - 2 q^2 - 4 p r) x^3 + x^4.
\end{aligned}
$$
As explained in the comments, because
$$
h(x)=(x-a^4)(x-b^4)(x-c^4)(x-d^4)
$$ the answer is
$$
h(-1)=p^4+2 p^2 r^2-4 p q^2 r+q^4+r^4.
$$
