Question

Prove that :

$\left(i\right){\left(\frac{x^a}{x^b}\right)}^{\frac{1}{ab}}{\left(\frac{x^b}{x^c}\right)}^{\frac{1}{bc}}{\left(\frac{x^c}{x^a}\right)}^{\frac{1}{ca}}=1\left(ii\right)\frac{1}{1+x^{a-b}}+\frac{1}{1+x^{b-a}}=1$

Answer 1

$\left(i\right){\left(\frac{x^a}{x^b}\right)}^{\frac{1}{ab}}{\left(\frac{x^b}{x^c}\right)}^{\frac{1}{bc}}{\left(\frac{x^c}{x^a}\right)}^{\frac{1}{ca}}=1\text{L.H.S.}={\left(\frac{x^a}{x^b}\right)}^{\frac{1}{ab}}{\left(\frac{x^b}{x^c}\right)}^{\frac{1}{bc}}{\left(\frac{x^c}{x^a}\right)}^{\frac{1}{ca}}{\left(x^{a-b}\right)}^{\frac{1}{ab}}{\left(x^{b-c}\right)}^{\frac{1}{bc}}{\left(x^{c-a}\right)}^{\frac{1}{ca}}=x^{\frac{a-b}{ab}}{x}^{\frac{b-c}{bc}}{x}^{\frac{c-a}{ca}}\{{\left(x^a\right)}^b=x^{ab}\}=x^{\frac{a-b}{ab}+\frac{b-c}{bc}+\frac{c-a}{ca}}=x^{\frac{ac-bc+ab-ac+bc-ab}{abc}}=x^0=1=\text{R.H.S.}(\because x^0=1)\left(ii\right)\frac{1}{1+x^{a-b}}+\frac{1}{1+x^{b-a}}=1\text{L.H.S.}=\frac{1}{1+x^{a-b}}+\frac{1}{1+x^{b-a}}=\frac{1}{x^{a-a}+x^{a-b}}+\frac{1}{x^{b-b}+x^{b-a}}=\frac{1}{x^a.x^{-a}+x^a.x^{-b}}+\frac{1}{x^b.x^{-b}+x^b.x^{-a}}=\frac{1}{x^a(x^{-a}+x^{-b})}+\frac{1}{x^b(x^{-b}+x^{-a})}=\frac{1}{(x^{-a}+x^{-b})}[\frac{1}{x^a}+\frac{1}{x^b}]=\frac{1}{x^{-a}+x^{-b}}[x^{-a}+x^{-b}]=1=\text{R.H.S.}$