Find $a^{b+c} b^{c+a} c^{a+b}$ if $\frac {\log a}{b-c}=\frac{\log b}{c-a}=\frac{\log c}{a-b}$ If $\frac {\log a}{b-c}=\frac{\log b}{c-a}=\frac{\log c}{a-b}$, then what would be the value of $a^{b+c} \cdot b^{c+a} \cdot c^{a+b}$?
I'm unable to proceed.
Let $ \frac {\log a}{b-c}=\frac{\log b}{c-a}=\frac{\log c}{a-b}=k$

$\Rightarrow a=e^{k(b-c)}, b=e^{k(c-a)},c=e^{k(a-b)}$. Then the required value is $ e^{k(b^2-c^2)}. e^{k(c^2-a^2)}. e^{k(a^2-b^2)}=1$
Let $$ \frac {\log a}{b-c}=\frac{\log b}{c-a}=\frac{\log c}{a-b}=k\tag{1}$$

From $(1)$, we get 

$$\log{a}=k(b-c)\tag{2}$$

$$\log{b}=k(c-a)\tag{3}$$

$$\log{c}=k(a-b)\tag{4}$$

Let $$l=a^{b+c} \cdot b^{c+a} \cdot c^{a+b}$$

Taking $\log$ on both sides, we get 
 $$\log{l}=(b+c) \cdot \log{a}+(c+a) \cdot \log{b}+(a+b) \cdot \log{c}\tag{5}$$

Now, using $(2),(3)$ and $(4)$ in $(5)$, we get 

$$\log{l}=k(b^2-c^2)+k(c^2-a^2)+k(a^2-b^2)=0\tag{6}$$

From $(6)$, we get $\boxed{l=e^0=1}$