※ 引述《diow1 (小玉)》之銘言：
: 2 3
: ω ω ω
: ω是 1 的 7次方根 , 求 _______ + ________ + __________ = ???
: 2 4 6
: 1+ ω 1+ ω 1+ ω
= 1/(w+1/w)+1/(w^2+1/w^2) + 1/(w^3+1/w^3)
w satisfies 1+z+..+z^6=0,
or z^3+1/z^3 + z^2+1/z^2 + z+1/z + 1 =0 .
Let t = z+ 1/z, then
t^3 - 3t + t^2 - 2 + t + 1 = 0
t^3 + t^2 - 2t - 1 =0.
thus, 1/(w+1/w)+1/(w^2+1/w^2) + 1/(w^3+1/w^3) = -2/1 = -2.
--
※ 發信站: 批踢踢實業坊(ptt.cc)
◆ From: 131.215.6.212