==> ${1 \over {\ln a.\ln b.\ln c}}[{(\ln a)^3} + {(\ln b)^3} + {(\ln c)^3} - 3\ln a.\ln b.\ln c] = 0$
==> ${(\ln a)^3} + {(\ln b)^3} + {(\ln c)^3} - 3\ln a.\ln b.\ln c = 0$
==> $\ln a + \ln b + \ln c = 0$
==> $\ln (abc) = ln 1$, $[{a^3} + {b^3} + {c^3} - 3abc = 0$
==> $a + b + c = 0]$,
$\therefore abc = 1$.
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$(A)$ $\tan ^2 x=\frac{2}{3}$ $(B)$ $\frac{\sin ^8 x}{8}+\frac{\cos ^8 x}{27}=\frac{1}{125}$
$(C)$ $\tan ^2 x=\frac{1}{3}$ $(D)$ $\frac{\sin ^8 x}{8}+\frac{\cos ^8 x}{27}=\frac{2}{125}$