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

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.
