If 111abcbccaab=111abca3b3c3, where a, b, c are distinct positive reals, then the possible values of abc is/are
181
163
127
118
We have
(a−b)(b−c)(c−a)=(a−b)(b−c)(c−a)(a+b+c)(a≠b≠c)
⇒ a+b+c=1
As a, b and c are positive.
Using AM > GM, we get
a+b+c3>(abc)13∴ abc<127