If x is a positive real number and exponents are rational numbers, then simplify the following :

$\Big(\dfrac{x^{a^2}}{x^{b^2}}\Big)^{\dfrac{1}{a + b}}\Big(\dfrac{x^{b^2}}{x^{c^2}}\Big)^{\dfrac{1}{b + c}}\Big(\dfrac{x^{c^2}}{x^{a^2}}\Big)^{\dfrac{1}{c + a}}$

Given,

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

Hence, $\Big(\dfrac{x^{a^2}}{x^{b^2}}\Big)^{\dfrac{1}{a + b}}\Big(\dfrac{x^{b^2}}{x^{c^2}}\Big)^{\dfrac{1}{b + c}}\Big(\dfrac{x^{c^2}}{x^{a^2}}\Big)^{\dfrac{1}{c + a}}$ = 1.