20/11/06 20:03:24.22 2uQNgYSq.net
αは1の6乗根とする。
α^6 =1, α^3≠1, α^2≠1, α≠-1
(α^2 -1)(α^4 +α^2 +1) = α^6 -1 = 0, α^2≠１
∴ α^4 + α^2 +1 = 0,
∴ (1-α^2)(1-α^4) = 3 - (α^4 +α^2 +1) = 3,
(α^3 -1)(α^3 +1) = α^6 -1 = 0, α^3≠1
∴ α^3 +1 = 0,
∴ (1-α^3) = 2,
(α+1)(α^2 -α +1) = α^3 +1 = 0, α≠-1
∴ α^2 -α +1 = 0,
∴ (1-α)(1-α^5) = (1-α)(1-1/α) = 1 - (α^2 - α+1)/α = 1,
辺々掛けて 3･2･1 = 6