Question

Prove that : ${\left(\frac{x^a}{x^b}\right)}^{\frac{1}{ab}}.{\left(\frac{x^b}{x^c}\right)}^{\frac{1}{bc}}.{\left(\frac{x^c}{x^a}\right)}^{\frac{1}{ca}}=1$

Answer 1

Consider LHS : ${\left(\frac{x^a}{x^b}\right)}^{\frac{1}{ab}}\times {\left(\frac{x^b}{x^c}\right)}^{\frac{1}{bc}}\times {\left(\frac{x^c}{x^a}\right)}^{\frac{1}{ac}}$

$={\left(x^{a-b}\right)}^{\frac{1}{ab}}\times {\left(x^{b-c}\right)}^{\frac{1}{bc}}\times {\left(x^{c-a}\right)}^{\frac{1}{ac}}[\because \frac{a^m}{a^n}=a^{m-n}]$

$=x^{\left(\frac{a-b}{ab}\right)}\times x^{\left(\frac{b-c}{bc}\right)}\times x^{\left(\frac{c-a}{ac}\right)}$

$=x^{\left(\frac{1}{b}-\frac{1}{a}\right)}\times x^{\left(\frac{1}{c}-\frac{1}{b}\right)}\times x^{\left(\frac{1}{a}-\frac{1}{c}\right)}$

$=x^{\left(\frac{1}{b}-\frac{1}{a}+\frac{1}{c}-\frac{1}{b}+\frac{1}{a}-\frac{1}{c}\right)}[\because a^m\times a^n=a^{m+n}]$

$=x^0=1$