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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?

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2 Answers

$$\begin{align} a+b+c+d &= \phantom{-}1 \\ ab+bc+cd+ac+ad+bd &= \phantom{-}2 \\ abc+bcd+abd+acd &= -1 \\ abcd &= -1 \end{align}$$

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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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.

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