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

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 $a^ab^bc^c=1$.

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.

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: $$a^ab^bc^c=e^{a(b-c)t}e^{b(c-a)t}e^{c(a-b)t}=e^{(ab-ac+bc-ab+ac-bc)t}=e^0=1$$