數學白板/YuLM/2008-08-26-15

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$1,\omega _{1},\omega _{2},\omega _{3},\omega _{4}\,$ 為方程式 $x^{5}=1\,$ 之 $5\,$ 個根

$\left( 3-\omega _{1} \right)\left( 3-\omega _{2} \right)\left( 3-\omega _{3} \right)\left( 3-\omega _{4} \right)=?$

令 $f\left( x \right)=x^{5}-1=\left( x-1 \right)\left( x^{4}+x^{3}+x^{2}+x+1 \right)$
$\therefore \left\{ \begin{align} & f\left( 1 \right)=0 \\ & f\left( \omega _{1} \right)=0 \\ & f\left( \omega _{2} \right)=0 \\ & f\left( \omega _{3} \right)=0 \\ & f\left( \omega _{4} \right)=0 \\ \end{align} \right.\Rightarrow f\left( x \right)=\left( x-1 \right)\left( x-\omega _{1} \right)\left( x-\omega _{2} \right)\left( x-\omega _{3} \right)\left( x-\omega _{4} \right)$
$\Rightarrow \left( x-1 \right)\left( x-\omega _{1} \right)\left( x-\omega _{2} \right)\left( x-\omega _{3} \right)\left( x-\omega _{4} \right)=\left( x-1 \right)\left( x^{4}+x^{3}+x^{2}+x+1 \right)$
$\Rightarrow \left\{ \begin{align} & f\left( 3 \right)=\left( 3-1 \right)\left( 3-\omega _{1} \right)\left( 3-\omega _{2} \right)\left( 3-\omega _{3} \right)\left( 3-\omega _{4} \right) \\ & f\left( 3 \right)=\left( 3-1 \right)\left( 3^{4}+3^{3}+3^{2}+3+1 \right) \\ \end{align} \right.$
$\Rightarrow \left( 3-\omega _{1} \right)\left( 3-\omega _{2} \right)\left( 3-\omega _{3} \right)\left( 3-\omega _{4} \right)=\left( 3^{4}+3^{3}+3^{2}+3+1 \right)$
$\Rightarrow \left( 3-\omega _{1} \right)\left( 3-\omega _{2} \right)\left( 3-\omega _{3} \right)\left( 3-\omega _{4} \right)=\left( 3^{4}+3^{3}+3^{2}+3+1 \right)=121\#$