Question

For a>0,b>0,c>0 let x=ln(a/b) , y=ln(b/c) and z=ln(c/a).
Prove that x^3+y^3+z^3-3xyz=0.

Answer 1

\(x=\ln{\frac{a}{b}}=\ln a-\ln b, y=\ln b-\ln c, z=\ln c-\ln a.\)
\(x^3+y^3+z^3-3xyz=(x+y+z)(x^2-xy-xz+y^2-yz+z^2)\). But \(x+y+z=\ln a-\ln b+\ln b-\ln c+\ln c-\ln a=0\), so \(x^3+y^3+z^3-3xyz=0\).