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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.