<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] Define $w_1=(1+\sqrt{3}i)/2$ and $w_2=(1-\sqrt{3}i)/2$  where $i=\sqrt{-1}$. Now observe that 
\begin{align}
(a^3+1)=(a+1)(a-w_1)(a-w_2)
\end{align}
Similarily for other terms also. This helps you to write
\begin{align}
(a^3+1)(b^3+1)(c^3+1)(d^3+1)=P_1P_2P_3
\end{align}
where 
\begin{align}
P_1&=(a+1)(b+1)(c+1)(d+1) \\
P_2&=(a-w1)(b-w_1)(c-w_1)(d-w_1) \\
P_3&=(a-w2)(b-w_2)(c-w_2)(d-w_2)
\end{align}
Now for any $x$, you can write your original quartic equation as 
\begin{align}
f(x)&=x^4-x^3+2x^2+x+1 \\
    &=(a-x)(b-x)(c-x)(d-x)
\end{align}
Put $x=-1$. You obtain
\begin{align}
f(-1)=(a+1)(b+1)(c+1)(d+1)=P_1
\end{align}
So you get $P_1=f(-1)=4$. Similarily $P_2=f(w1)=1+\sqrt{3}i$ $P_3=f(w2)=1-\sqrt{3}i$ and hence $P_1P_2P_3=4*(1+\sqrt{3}i)*(1-\sqrt{3}i)=16$ and your answer. 
</s>