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)$