If $\cos(A+B+C)=\cos A\cos B\cos C\neq 0$, then evaluate $\left|\frac{8\sin(A+B)\sin(B+C)\sin(C+A)}{\sin 2A\sin 2B\sin 2C}\right|$ 
If 
  $$\cos(A+B+C)=\cos A\cos B\cos C,  \quad\text{with}\;A,B,C\neq \frac{k\pi}{2}$$
  then 
  $$\left|\frac{8\sin(A+B)\sin(B+C)\sin(C+A)}{\sin 2A\sin 2B\sin 2C}\right|$$ is what integer?

What I tried:
$$\begin{align}
\cos(A+B+C)
&=\cos A \cos B \cos C - \sin A \sin B \sin C - \sin C \sin A \cos B - \sin B \sin C \cos A \\
&=\cos A\cos B\cos C\left(1-\tan A\tan B-\tan B\tan C-\tan C\tan A\right)
\end{align}$$
How do i solve it Help me please 

Using $\sin(2x)=2\sin(x)\cos(x)$ along with $\sin(A+B)=\sin(\pi-C)=\sin(C)$ and $\cos(A)\cos(B)\cos(C)=\cos(A+B+C)=\cos(\pi)=-1$ should give the value as $1$.
