Question

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

Answer 1

Proof:

Consider the LHS of the given expression.

$$\begin{gathered} \mathrm{LHS}={\left(\frac{x^a}{x^b}\right)}^{1/ab}·{\left(\frac{x^b}{x^c}\right)}^{1/bc}·{\left(\frac{x^c}{x^a}\right)}^{1/ca} \\ \Rightarrow =x^{\frac{a-b}{ab}}·x^{\frac{b-c}{bc}}·x^{\frac{c-a}{ca}}\mathbf{}\mathbf{[}\mathbf{\because }\mathbf{U}\mathbf{s}\mathbf{e}\mathbf{}\frac{{\mathbf{a}}^{\mathbf{m}}}{{\mathbf{a}}^{\mathbf{n}}}\mathbf{=}{\mathbf{a}}^{\mathbf{m}\mathbf{-}\mathbf{n}}\mathbf{}\mathbf{}\mathbf{}\mathbf{a}\mathbf{n}\mathbf{d}\mathbf{}\mathbf{}{\mathbf{\left(}{\mathbf{a}}^{\mathbf{m}}\mathbf{\right)}}^{\mathbf{n}}\mathbf{=}{\mathbf{a}}^{\mathbf{m}\mathbf{n}}\mathbf{]} \\ \Rightarrow ={{x^{\frac{a-b}{ab}+}}^{\frac{b-c}{bc}+}}^{\frac{c-a}{ca}} \\ \Rightarrow =x^{\frac{c(a-b)+a(b-c)+b(c-a)}{abc}} \\ \Rightarrow =x^{\frac{ac-bc+ab-ac+bc-ab}{abc}} \\ \Rightarrow =x^{\frac{0}{abc}} \\ \Rightarrow =x^0\mathbf{[}\mathbf{\because }\mathbf{}\mathbf{}{\mathbf{a}}^{\mathbf{0}}\mathbf{=}\mathbf{1}\mathbf{]} \\ \Rightarrow =1 \\ \Rightarrow LHS=\mathrm{RHS} \end{gathered}$$

Hence proved.