==> \({1 \over {\ln a.\ln b.\ln c}}[{(\ln a)^3} + {(\ln b)^3} + {(\ln c)^3} - 3\ln a.\ln b.\ln c] = 0\)
==> \({(\ln a)^3} + {(\ln b)^3} + {(\ln c)^3} - 3\ln a.\ln b.\ln c = 0\)
==> \(\ln a + \ln b + \ln c = 0\)
==> \(\ln (abc) = ln 1\), \([{a^3} + {b^3} + {c^3} - 3abc = 0\)
==> \(a + b + c = 0]\),
\(\therefore abc = 1\).