<s>[INST] 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 
 [/INST] So, where you have left off,  
$$pq+qr+rp=0$$   writing  $\tan A=p,\tan B=q,\tan C=r$ 
we have $p,q,r$ are non-zero and finite
Now $$\dfrac{2\sin(A+B)}{\sin2C}=\cdots=\dfrac{p+q}{r}$$
$$\implies\dfrac{2\sin(A+B)\cdot2\sin(B+C)\cdot2\sin(C+A)}{\sin2A\sin2B\sin2C}$$
$$=\dfrac{(p+q)(q+r)(r+p)}{pqr}$$
$$=\dfrac{2pqr+p(pq+qr+rp)-pqr+q(pq+qr+rp)-pqr+r(pq+qr+rp)-pqr}{pqr}$$
$$=\dfrac{pqr(2-3)}{pqr}=?$$
</s>