Logical Logarithms

log a = m (b - c)

log b = m(c - a)

log c = m(a - b)

log (a^a)(b^b)(c^c) = a log a + b log b + c log c =

a m (b - c) + b m (c - a) + c m (a - b) = 0

Then

(a^a)(b^b)(c^c) = 1

Assume $\large{ \frac{\log a}{b-c}} =\frac{\log b}{c- a} =\frac{\log c}{a-b} = k \quad (k \in \Re)$

Then,

$\large \displaystyle \log a = k\cdot b -k\cdot c \implies a\log a = k(a\cdot b - a\cdot c)$

$\large\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad \implies \log a^{a} = k(a\cdot b - a\cdot c)$

Similarly, $\displaystyle \log b^{b} = k(c\cdot b - a\cdot b), \log c^{c} = k(a\cdot c - b\cdot c)$

Therefore,by adding the above results,

$\displaystyle \large \log \left(a^a\cdot b^b\cdot c^c\right) = k(a\cdot b - a\cdot c) + k(c\cdot b - a\cdot b) + k(a\cdot c - b\cdot c)$ $\displaystyle \large \quad \quad\quad\quad\quad\quad\quad\quad = 0$

Hence, $\displaystyle \large a^a\cdot b^b\cdot c^c = \boxed{1}$

I have used a cheat code to solve this question. Since we are equating three terms, we can assume that these three expressions have limits when all a, b, c are almost equal (i.e., c -> a and b -> a). But in that case, the denominator would be too small. In order for the expression to have a limit, the numerator should also be infinitesimally small. In other words, the expression should be in 0/0 form.

This is possible only if log a = 0 or a = 1. Since a,b,c are almost equal, we can write a=b=c=1

Therefore a^a * b^b * c^c = 1^1 1^1 1^1 = 1 Therefore the answer is $\boxed{1}$