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Question

# Prove that: (i) $\frac{1}{1+{x}^{a-b}}+\frac{1}{1+{x}^{b-a}}=1$ (ii) $\frac{1}{1+{x}^{b-a}+{x}^{c-a}}+\frac{1}{1+{x}^{a-b}+{x}^{c-b}}+\frac{1}{1+{x}^{b-c}+{x}^{a-c}}=1$

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Solution

## (i) Consider the left hand side: $\frac{1}{1+{x}^{a-b}}+\frac{1}{1+{x}^{b-a}}\phantom{\rule{0ex}{0ex}}=\frac{1}{1+\frac{{x}^{a}}{{x}^{b}}}+\frac{1}{1+\frac{{x}^{b}}{{x}^{a}}}\phantom{\rule{0ex}{0ex}}=\frac{1}{\frac{{x}^{b}+{x}^{a}}{{x}^{b}}}+\frac{1}{\frac{{x}^{a}+{x}^{b}}{{x}^{a}}}\phantom{\rule{0ex}{0ex}}=\frac{{x}^{b}}{{x}^{b}+{x}^{a}}+\frac{{x}^{a}}{{x}^{a}+{x}^{b}}\phantom{\rule{0ex}{0ex}}=\frac{{x}^{b}+{x}^{a}}{{x}^{b}+{x}^{a}}\phantom{\rule{0ex}{0ex}}=1$ Therefore left hand side is equal to the right hand side. Hence proved. (ii) Consider the left hand side: $\frac{1}{1+{x}^{b-a}+{x}^{c-a}}+\frac{1}{1+{x}^{a-b}+{x}^{c-b}}+\frac{1}{1+{x}^{b-c}+{x}^{a-c}}\phantom{\rule{0ex}{0ex}}=\frac{1}{1+\frac{{x}^{b}}{{x}^{a}}+\frac{{x}^{c}}{{x}^{a}}}+\frac{1}{1+\frac{{x}^{a}}{{x}^{b}}+\frac{{x}^{c}}{{x}^{b}}}+\frac{1}{1+\frac{{x}^{b}}{{x}^{c}}+\frac{{x}^{a}}{{x}^{c}}}\phantom{\rule{0ex}{0ex}}=\frac{1}{\frac{{x}^{a}+{x}^{b}+{x}^{c}}{{x}^{a}}}+\frac{1}{\frac{{x}^{b}+{x}^{a}+{x}^{c}}{{x}^{b}}}+\frac{1}{\frac{{x}^{c}+{x}^{b}+{x}^{c}}{{x}^{c}}}\phantom{\rule{0ex}{0ex}}=\frac{{x}^{a}}{{x}^{a}+{x}^{b}+{x}^{c}}+\frac{{x}^{b}}{{x}^{b}+{x}^{a}+{x}^{c}}+\frac{{x}^{c}}{{x}^{c}+{x}^{b}+{x}^{c}}\phantom{\rule{0ex}{0ex}}=\frac{{x}^{a}+{x}^{b}+{x}^{c}}{{x}^{a}+{x}^{b}+{x}^{c}}\phantom{\rule{0ex}{0ex}}=1$ Therefore left hand side is equal to the right hand side. Hence proved.

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