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Question Number 38477 by Sr@2004 last updated on 26/Jun/18

Answered by $@ty@m last updated on 26/Jun/18

$$\frac{a-b}{c}+\frac{b-c}{a}+\frac{c+a}{b}=1$$$$\Rightarrow \frac{a-b}{c}+\frac{c+a}{b}=1-\frac{b-c}{a}$$$$\Rightarrow \frac{b(a-b)+c(c+a)}{bc}=\frac{a-b+c}{a}$$$$\Rightarrow \frac{(a-b+c)(b+c)}{bc}=\frac{a-b+c}{a}$$$$\Rightarrow \frac{b+c}{bc}=\frac{1}{a}$$$$\Rightarrow \frac{1}{b}+\frac{1}{c}=\frac{1}{a}$$

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