Logs

Let $\dfrac {\log a}{b-c} = \dfrac {\log b}{c-a} = \dfrac {\log c}{a-b} = \color{#3D99F6}k$ .

$\implies \begin{cases} \log a = k(b-c) & \implies a = e^{k(b-c)} \\ \log b = k(c-a) & \implies b = e^{k(c-a)} \\ \log c = k(a-b) & \implies c = e^{k(a-b)} \end{cases}$

$\implies a^ab^bc^c = e^{ak(b-c)}e^{bk(c-a)}e^{ck(a-b)} = e^{k(ab-ca+bc-ab+ca-bc)} = e^0 = \boxed{1}$