Prove that:

$\Big(\dfrac{x^a}{x^b}\Big)^{\dfrac{1}{ab}} \times \Big(\dfrac{x^b}{x^c}\Big)^{\dfrac{1}{bc}} \times \Big(\dfrac{x^c}{x^a}\Big)^{\dfrac{1}{ac}} = 1$

Given,

$\Big(\dfrac{x^a}{x^b}\Big)^{\dfrac{1}{ab}} \times \Big(\dfrac{x^b}{x^c}\Big)^{\dfrac{1}{bc}} \times \Big(\dfrac{x^c}{x^a}\Big)^{\dfrac{1}{ac}} = 1$

Solving L.H.S :

$\Rightarrow \Big(\dfrac{x^a}{x^b}\Big)^{\dfrac{1}{ab}} \times \Big(\dfrac{x^b}{x^c}\Big)^{\dfrac{1}{bc}} \times \Big(\dfrac{x^c}{x^a}\Big)^{\dfrac{1}{ac}} \\[1em] \Rightarrow (x^{a - b})^{\dfrac{1}{ab}} \times (x^{b - c})^{\dfrac{1}{bc}} \times (x^{c - a})^{\dfrac{1}{ca}} \\[1em] \Rightarrow (x)^{\dfrac{a - b}{ab}} \times (x)^{\dfrac{b - c}{bc}} \times (x)^{\dfrac{c - a}{ac}} \\[1em] \Rightarrow (x)^{\dfrac{a}{ab}- \dfrac{b}{ab}} \times (x)^{\dfrac{b}{bc} -\dfrac{c}{bc}} \times (x)^{\dfrac{c}{ac}- \dfrac{a}{ac}} \\[1em] \Rightarrow \Big[(x)^{\dfrac{1}{b} - \dfrac{1}{a}}\Big] \times \Big[(x)^{\dfrac{1}{c} - \dfrac{1}{b}}\Big] \times \Big[(x)^{\dfrac{1}{a} - \dfrac{1}{c}}\Big] \\[1em] \Rightarrow \Big[(x)^{\dfrac{1}{b} - \dfrac{1}{a} + \dfrac{1}{c} - \dfrac{1}{b} + \dfrac{1}{a} - \dfrac{1}{c}}\Big] \\[1em] \Rightarrow x^0 \\[1em] \Rightarrow 1.$

Hence proved, $\Big(\dfrac{x^a}{x^b}\Big)^{ab} \times \Big(\dfrac{x^b}{x^c}\Big)^{bc} \times \Big(\dfrac{x^c}{x^a}\Big)^{ca} = 1$.