# Prove that:

Question:

Prove that:

(i) $\frac{1}{1+x^{a-b}}+\frac{1}{1+x^{b-a}}=1$

(ii) $\frac{1}{1+x^{b-a}+x^{c-a}}+\frac{1}{1+x^{a-b}+x^{c-b}}+\frac{1}{1+x^{b-c}+x^{a-c}}=1$

Solution:

(i) Consider the left hand side:

$\frac{1}{1+x^{a-b}}+\frac{1}{1+x^{b-a}}$

$=\frac{1}{1+\frac{x^{a}}{x^{b}}}+\frac{1}{1+\frac{x^{b}}{x^{a}}}$

$=\frac{1}{\frac{x^{b}+x^{a}}{x^{b}}}+\frac{1}{\frac{x^{a}+x^{b}}{x^{a}}}$

$=\frac{x^{b}}{x^{b}+x^{a}}+\frac{x^{a}}{x^{a}+x^{b}}$

$=\frac{x^{b}+x^{a}}{x^{b}+x^{a}}$

$=1$

Therefore left hand side is equal to the right hand side. Hence proved.

(ii)

Consider the left hand side:

$\frac{1}{1+x^{b-a}+x^{c-a}}+\frac{1}{1+x^{a-b}+x^{c-b}}+\frac{1}{1+x^{b-c}+x^{a-c}}$

$=\frac{1}{1+\frac{x^{b}}{x^{a}}+\frac{x^{c}}{x^{a}}}+\frac{1}{1+\frac{x^{a}}{x^{b}}+\frac{x^{c}}{x^{b}}}+\frac{1}{1+\frac{x^{b}}{x^{c}}+\frac{x^{a}}{x^{c}}}$

$=\frac{1}{\frac{x^{a}+x^{b}+x^{c}}{x^{a}}}+\frac{1}{\frac{x^{b}+x^{a}+x^{c}}{x^{b}}}+\frac{1}{\frac{x^{c}+x^{b}+x^{c}}{x^{c}}}$

$=\frac{x^{a}}{x^{a}+x^{b}+x^{c}}+\frac{x^{b}}{x^{b}+x^{a}+x^{c}}+\frac{x^{c}}{x^{c}+x^{b}+x^{c}}$

$=\frac{x^{a}+x^{b}+x^{c}}{x^{a}+x^{b}+x^{c}}$

$=1$

Therefore left hand side is equal to the right hand side. Hence proved.