$\begin{align} & \text{By putting},\alpha =\log a,\text{ }\beta =\log b,\text{ }\gamma =\log c\text{ in the identity } \\ & \text{show that }\alpha (\beta -\gamma )+\beta (\gamma -\alpha )+\gamma (\alpha -\beta )=0 \\\end{align}$
$\begin{align} & \log a(\log b-\log c)+\log b(\log c-\log a)+\log c(\log a-\log b)=0 \\ & \log a\left( \log \tfrac{b}{c} \right)+\log b\left( \log \tfrac{c}{a} \right)+\log c\left( \log \tfrac{a}{b} \right)=0 \\ & \log {{\tfrac{b}{c}}^{\log a}}+\log {{\tfrac{c}{a}}^{\log a}}+\log {{\tfrac{a}{b}}^{\log c}}=0 \\ & \log \left( {{\tfrac{b}{c}}^{\log a}}\times {{\tfrac{c}{a}}^{\log a}}\times {{\tfrac{a}{b}}^{\log c}} \right)=\log 1 \\ & {{\tfrac{b}{c}}^{\log a}}\times {{\tfrac{c}{a}}^{\log a}}\times {{\tfrac{a}{b}}^{\log c}}=1 \\\end{align}$