推 linkismet :似乎的確是2，下面是我的計算 06/23 04:02
已知x,y為非零複數,且 x^2 + xy + y^2 = 0,
求(x/(x + y))^102 + (y/(x + y))^102 = ?
聽說答案是 -1 , 但是我用底下2個解法算出來都是 2
麻煩各路高手幫忙看一下問題出在哪?
問題解1 : (x-y)(x^2 + xy + y^2) = 0 (條件同乘 x-y)
可得 x^3 - y^3 = 0 , 移項 x^3 = y^3
條件同加 xy x^2 + 2xy + y^2 = xy , 可得 (x + y)^2 = xy
(x/(x + y))^102 + (y/(x + y))^102 = (x^102 + y^102)/(x + y)^102
= ( (x^3)^34 + (y^3)^34 ) / ( (x + y)^2 )^51
= ( (y^3)^34 + (y^3)^34 ) / (xy)^51
= 2 * y^102 / ( (x^3)^17 * (y^3)^17 )
= 2 * y^102 / ( (y^3)^17 * (y^3)^17 ) = 2 * y^102 / y^102 = 2
問題解2 : (x-y)(x^2 + xy + y^2) = 0 (條件同乘 x-y)
可得 x^3 - y^3 = 0 , 移項 x^3 = y^3
(x/(x + y))^102 + (y/(x + y))^102 = (x^102 + y^102)/(x + y)^102
= ( (x^3)^34 + (y^3)^34 ) / (x + y)^102
= ( (y^3)^34 + (y^3)^34 ) / (x + y)^102
= 2 / (x/y + 1)^102
由上面條件同除以 y^3 (x/y)^3 = 1
令 x/y = a, a^3 = 1 的解 = 1, -0.5 ±√3/2 i
若 x/y = 1 則 x = y, x^2 + xy + y^2 = 3 x^2 = 0 將得到 x = y = 0 (不合)
所以 x/y + 1 = 0.5 ±√3/2 i 是 六次方程式 ω^6 = 1 的解
2 / (x/y + 1)^102 = 2 / ( (ω^6)^17 ) = 2
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