<s>[INST] If $a,b,c,d$ be the roots of the biquadratic $x^4-x^3+2x^2+x+1=0$ then show that $(a^3+1)(b^3+1)(c^3+1) (d^3+1)=16$ If $a,b,c,d$ be the roots of the biquadratic $x^4-x^3+2x^2+x+1=0$ then show that  $(a^3+1)(b^3+1)(c^3+1) (d^3+1)=16$
I have tried to solve the equation first and find the values of the roots but it becomes very long process. Is there any easy process?
 [/INST] $$\begin{align}
a+b+c+d &= \phantom{-}1 \\
ab+bc+cd+ac+ad+bd &= \phantom{-}2 \\
abc+bcd+abd+acd &= -1 \\
abcd &= -1
\end{align}$$
All you need to find is
$$\begin{align}
(abc)^3 + (abd)^3 + (acd)^3 + (bcd)^3 &= p \\
(ab)^3 + (ac)^3 + (ad)^3 + (bc)^3 + (bd)^3 + (cd)^3 & = q \\
a^3 + b^3 + c^3 +d^3 &= r
\end{align}$$
And you need to show
$$(abcd)^3 + p + q + r + 1 = 16$$
which is really just
$$p+q+r = 16$$
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