Question

Show that

${\left(\frac{x^{a^2}}{x^{b^2}}\right)}^{\frac{1}{a+b}}\times {\left(\frac{x^{b^2}}{x^{c^2}}\right)}^{\frac{1}{b+c}}\times {\left(\frac{x^{c^2}}{x^{a^2}}\right)}^{\frac{1}{c+a}}=1$

Answer 1

$\begin{array}{c}LHS={\left(\frac{x^{a^2}}{x^{b^2}}\right)}^{\frac{1}{a+b}}\times {\left(\frac{x^{b^2}}{x^{c^2}}\right)}^{\frac{1}{b+c}}\times {\left(\frac{x^{c^2}}{x^{a^2}}\right)}^{\frac{1}{c+a}}\\ \left[since,\frac{x^m}{x^n}=x^{m-n}\right]\\ ={\left(x^{a^2-b^2}\right)}^{\frac{1}{a+b}}\times {\left(x^{b^2-c^2}\right)}^{\frac{1}{b+c}}\times {\left(x^{c^2-a^2}\right)}^{\frac{1}{c+a}}\\ ={\left(x\right)}^{\frac{a^2-b^2}{a+b}}\times {\left(x\right)}^{\frac{b^2-c^2}{b+c}}\times {\left(x\right)}^{\frac{c^2-a^2}{c+a}}\\ ={\left(x\right)}^{\frac{\left(a-b\right)\left(a+b\right)}{\left(a+b\right)}}\times {\left(x\right)}^{\frac{\left(b-c\right)\left(b+c\right)}{\left(b+c\right)}}\times {\left(x\right)}^{\frac{\left(c+a\right)\left(c-a\right)}{\left(c+a\right)}}\\ ={\left(x\right)}^{\left(a-b\right)}\times {\left(x\right)}^{\left(b-c\right)}\times {\left(x\right)}^{\left(c-a\right)}\\ ={\left(x\right)}^{\left(a-b\right)+\left(b-c\right)+(c-a)}\\ =1=RHS\end{array}$