Question 1040182
Let {{{loga/(b-c) = logb/(c-a) = logc/(a-b) = k}}}.

==> {{{a = 10^(k(b-c))}}},
{{{b = 10^(k(c-a))}}}, and 
{{{c = 10^(k(a-b))}}}.

==> {{{abc = 10^(k(b-c))*10^(k(c-a))*10^(k(a-b)) = 10^(kb-kc+kc-ka+ka-kb) = 10^0 = 1}}}.