<s>[INST] 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
 [/INST] 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
$$a^a.b^b.c^c=1.$$
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