Докажите тождество:
$\frac{1}{(b - c)(c - a)} - \frac{1}{(a - b)(c - b)} + \frac{1}{(a - c)(b - a)} = 0$.
$\frac{1}{(b - c)(c - a)} - \frac{1}{(a - b)(c - b)} + \frac{1}{(a - c)(b - a)} = -\frac{1}{(b - c)(a - c)} + \frac{1}{(a - b)(b - c)} - \frac{1}{(a - c)(a - b)} = \frac{-(a - b) + a - c - (b - c)}{(a - b)(b - c)(a - c)} = \frac{-a + b + a - c - b + c}{(a - b)(b - c)(a - c)} = \frac{0}{(a - b)(b - c)(a - c)} = 0$
Пожауйста, оцените решение
