Question

Доказать, что если:

x = a-b/a+b; y = b-c/b+c; z = c-a/c+a,

то (1 + x)(1 + y)(1 + z) = (1 - x)(1 - y)(1 - z)

Answer 1

$(1 + x)(1 + y)(1 + z) = \\ = (1 + \frac{a - b}{a + b} )(1 + \frac{b - c}{b + c})(1 + \frac{c - a}{c + a} ) = \\ = \frac{a + b + a - b}{a + b} \times \frac{b + c + b - c}{b + c} \times \\ \frac{c + a + c - a}{c + a } = \frac{2a \times 2b \times 2c}{(a +b )(b + c)(c + a)} = \\ = \frac{8abc}{(a +b )(b + c)(c + a)}$
$(1 - x)(1 - y)(1 - z) = \\ = (1 - \frac{a - b}{a + b} )(1 - \frac{b - c}{b + c})(1 - \frac{c - a}{c + a} ) = \\ = \frac{a + b - ( a - b)}{a + b} \times \frac{b + c - ( b - c)}{b + c} \times \\ \frac{c + a - ( c - a)}{c + a } = \frac{2b \times 2c \times 2a}{(a +b )(b + c)(c + a)} = \\ = \frac{8abc}{(a +b )(b + c)(c + a)}$
Отсюда следует, что

$(1 + x)(1 + y)(1 + z) = \\ = (1 - x)(1 - y)(1 - z)$

Answer 2

$(1+x)(1+y)(1+z)=(1+\frac{a-b}{a+b})(1+\frac{b-c}{b+c})(1+\frac{c-a}{c+a})=\\\\=\frac{a+b+a-b}{a+b}\cdot \frac{b+c+b-c}{b+c}\cdot \frac{c+a+c-a}{c+a}=\frac{2a}{a+b}\cdot \frac{2b}{b+c}\cdot \frac{2c}{c+a}=\frac{8\, abc}{(a+b)(b+c)(c+a)}\\\\\\(1-x)(1-y)(1-z)=(1-\frac{a-b}{a+b})(1-\frac{b-c}{b+c})(1-\frac{c-a}{c+a})=\\\\=\frac{a+b-a+b}{a+b}\cdot \frac{b+c-b+c}{b+c}\cdot \frac{c+a-c+a}{c+a}=\frac{2b}{a+b}\cdot \frac{2c}{b+c}\cdot \frac{2a}c+a}=\frac{8\, abc}{(a+b)(b+c)(c+a)}\\\\\\\Rightarrow \; \; \; (1+x)(1+y)(1+z)=(1-x)(1-y)(1-z)$