Юра Мальцев
Гуру
(2821)
2 дня назад
([a, b][b, c][c, a]) = (a × b) ⋅ (b × c) × (c × a) = (b × c) ⋅ (c × a) × (a × b) = [(b × c) × (c × a)] ⋅ a = = [(b ⋅ a)c - (c ⋅ a)b] ⋅ a = (b ⋅ a)(c ⋅ a) - (c ⋅ a)(b ⋅ a) = 0
(a, b, c)^2 = (a, b, c)(a, b, c) = a ⋅ (b × c) * a ⋅ (b × c) = (a × b) ⋅ (b × c) × (c × a) = 0
0 = 0 Ч Т Д
Роман РуссоЗнаток (369)
2 дня назад
a ⋅ (b × c) * a ⋅ (b × c) = (a × b) ⋅ (b × c) × (c × a)
Не могли бы Вы ещё подсказать, откуда следует этот переход?