Prove that$ \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 $

Given:

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

To do:

We have to prove that \( \frac{1}{1+x^{b-a}+x^{c-a}}+\frac{1}{1+x^{a-b}+x^{c-b}}+\frac{1}{1+x^{b-c}+x^{a-c}}=1 \).

Solution:

We know that,

$(a^{m})^{n}=a^{m n}$

$a^{m} \times a^{n}=a^{m+n}$

$a^{m} \div a^{n}=a^{m-n}$

$a^{0}=1$  

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

$=\frac{1}{x^{a-a}+x^{b-a}+x^{c-a}}+\frac{1}{x^{b-b}+x^{a-b}+x^{c-b}}+\frac{1}{x^{c-c}+x^{b-c}+x^{a-c}}$           [Substitute $1=x^{a-a}, 1=x^{b-b}$ and $1=x^{c-c}$]

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

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

$=1$

$=$ RHS

Hence proved.