# Prove that:

Question:

Prove that:

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

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

Solution:

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

Left hand side (LHS) = Right hand side (RHS) Considering LHS,

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

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

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

$=1$

Therefore, LHS = RHS Hence proved

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

Left hand side (LHS) = Right hand side (RHS) Considering LHS,

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

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

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

$=1$

Therefore, LHS = RHS Hence proved