Prove that:

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

Given,

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

Solving L.H.S :

$\Rightarrow \Big(\dfrac{x^a}{x^b}\Big)^{a + b} \times \Big(\dfrac{x^b}{x^c}\Big)^{b + c} \times \Big(\dfrac{x^c}{x^a}\Big)^{c + a} \\[1em] \Rightarrow (x^{a - b})^{a + b} \times (x^{b - c})^{b + c} \times (x^{c - a})^{c + a} \\[1em] \Rightarrow (x^{a^2 - b^2}) \times (x^{b^2 - c^2}) \times (x^{c^2 - a^2}) \\[1em] \Rightarrow (x^{a^2 - b^2 + b^2 - c^2 + c^2 - a^2}) \\[1em] \Rightarrow x^0 \\[1em] \Rightarrow 1.$

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