>>572

X = x-aω,　X~ = x-aω~,　ω≠1 は1の3乗根
とおくと
XX~ = xx+ax+aa,　(X-X~)^2 = -3aa,

∴（X^5 - 3)(X~^5 - 3)
= (XX~)^5 - 3[(X)^5 + (X~)^5] + 9
= (XX~)^5 - 3(X+X~)[(X^4) -(X^3)(X~) +(XX~)^2 -X(X~)^3 +(X~)^4] + 9
= (XX~)^5 - 3(X+X~)[(XX~)^2 + 3(XX~)(X-X~)^2 + (X-X~)^4] + 9
= (xx+ax+aa)^5 - 3(2x+a)[(xx+ax+aa)^2 - 9aa(xx+ax+aa) + 9a^4] + 9,

なぜ整数係数になるんでしょう？？？