If $\frac{\log a}{b-c}=\frac{\log b}{c-a}=\frac{\log c}{a-b}$ show that $a^a \cdot b^b\cdot c^c=1$ . 
If $\frac{\log a}{b-c}=\frac{\log b}{c-a}=\frac{\log c}{a-b}$ show that $a^a \cdot b^b\cdot c^c=1$ .

My Working:
$\frac{\log a}{b-c}= \frac{\log b}{c-a}$
$ (c-a)\log a=(b-c) \log b$
$ \log a^{c-a}=\log b^{b-c}$
$ \frac {a^c}{a^a}=\frac{b^b}{b^c}$
$ \frac {a^c \cdot{b^c}}{a^a} =b^b\qquad \text{(i)}$
Similarly, taking the next two terms we obtain,
$b^b=\frac{b^a \cdot c^a}{c^c}\qquad \text{(ii)}$
I tried to solve the two equations obtained to get to the desired statement but I couldn't. Is the way adopted correct or is there another way to reach the desired answer. Please help me proceed with this question
 A: Call $k$ the common value of
$$\frac{\log a}{b-c}=\frac{\log b}{c-a}=\frac{\log c}{a-b}$$
and then
$$\log(a^ab^bc^c)=a\log a+b\log b+c\log c=\left(a(b-c)+b(c-a)+c(a-b)\right)\cdot k=0$$
which implies that $a^ab^bc^c=1$.
A: It is given that $$\frac{\log a}{b-c}=\frac{\log b}{c-a}=\frac{\log c}{a-b}\tag1$$
Now $$\frac{\log a}{b-c}=\frac{\log b}{c-a}=\frac{\log a+\log b}{b-c+c-a} \,\,\,\,\,\,\,\text {(by Addendo)}$$
$$=\frac{\log ab}{b-a} \tag2$$
So from $(1)$ and $(2)$, we get that  $$\frac{\log c}{a-b}=\frac{\log ab}{b-a}$$
$$\implies ab =\frac {1}{c}\tag3$$
From equation $(i)$ established by you in the question and $(3) $,  we get $$\frac {a^c \cdot{b^c}}{a^a} =b^b$$
$$\implies  a^c \cdot b^c=a^a \cdot b^b$$
$$\implies  \frac {1}{c^c}=a^a \cdot b^b$$
$$\implies  a^a \cdot b^b \cdot c^c= 1$$
Hope this helps you.
A: Let $\frac{\log a}{b-c}=\frac{\log b}{c-a}=\frac{\log c}{a-b}=t\ $ implies $a=e^{(b-c)t}$, $b=e^{(c-a)t}$ & $c=e^{(a-b)t}$
so now, using values of a, b, c: $$a^ab^bc^c=e^{a(b-c)t}e^{b(c-a)t}e^{c(a-b)t}=e^{(ab-ac+bc-ab+ac-bc)t}=e^0=1$$
A: We use following property
if $F=\frac{A_1}{B_1}=\frac{A_2}{B_2}$
then 
$$F=\frac{A_1+A_2}{B_1+B_2}.$$
So, your equality implies
$$\frac{\ln(a^a)}{ab-ac}=\frac{\ln(b^b)}{bc-ba}=\frac{\ln(c^c)}{ca-cb}$$
$$=\frac{\ln(a^a)+\ln(b^b)}{cb-ca}$$
thus
$$a^a.b^b.c^c=1.$$
