Question

Prove that:

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

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

Answer 1

(i) Consider the left hand side:
$$\begin{gathered} \frac{1}{1+x^{a-b}}+\frac{{\displaystyle 1}}{{\displaystyle 1+x^{b-a}}} \\ =\frac{1}{1+{\displaystyle \frac{x^a}{x^b}}}+\frac{{\displaystyle 1}}{{\displaystyle 1+\frac{{\displaystyle x^b}}{x^a}}} \\ =\frac{1}{{\displaystyle \frac{x^b+x^a}{x^b}}}+\frac{{\displaystyle 1}}{{\displaystyle \frac{{\displaystyle x^a+x^b}}{x^a}}} \\ =\frac{x^b}{{\displaystyle x^b+x^a}}+\frac{{\displaystyle x^a}}{{\displaystyle x^a+x^b}} \\ =\frac{x^b+x^a}{{\displaystyle x^b+x^a}} \\ =1 \end{gathered}$$
Therefore left hand side is equal to the right hand side. Hence proved.

(ii)
Consider the left hand side:
$$\begin{gathered} \frac{{\displaystyle 1}}{{\displaystyle 1+x^{b-a}+x^{c-a}}}+\frac{1}{1+x^{a-b}+x^{c-b}}+\frac{1}{1+x^{b-c}+x^{a-c}} \\ =\frac{{\displaystyle 1}}{{\displaystyle 1+\frac{{\displaystyle x^b}}{x^a}+\frac{{\displaystyle x^c}}{x^a}}}+\frac{1}{1+{\displaystyle \frac{x^a}{x^b}}+{\displaystyle \frac{x^c}{x^b}}}+\frac{1}{1+{\displaystyle \frac{x^b}{x^c}}+{\displaystyle \frac{x^a}{x^c}}} \\ =\frac{{\displaystyle 1}}{{\displaystyle \frac{{\displaystyle x^a+x^b+x^c}}{x^a}}}+\frac{1}{{\displaystyle \frac{x^b+x^a+x^c}{x^b}}}+\frac{1}{{\displaystyle \frac{x^c+x^b+x^c}{x^c}}} \\ =\frac{{\displaystyle x^a}}{{\displaystyle x^a+x^b+x^c}}+\frac{x^b}{{\displaystyle x^b+x^a+x^c}}+\frac{x^c}{{\displaystyle x^c+x^b+x^c}} \\ =\frac{x^a+x^b+x^c}{{\displaystyle x^a+x^b+x^c}} \\ =1 \end{gathered}$$
Therefore left hand side is equal to the right hand side. Hence proved.