If $x+y+z=xyz$ then $\sum \frac{3x-x^3}{1-3x^2}=$

$\begin{array}{l}Put x=\tan A,y=\tan B,z=\tan C then x+y+z=xyz becomes\\ \tan A+\tan B+\tan C=\tan A\tan B\tan C\\ \Rightarrow A+B+C={180}^{\circ }\end{array}$
$\begin{array}{l}3A+3B+3C={540}^{\circ }\\ 3A+3B={540}^{\circ }-3C\\ \tan 3A+\tan 3B+\tan 3C=\tan 3A\tan 3B\tan 3C\\ \sum \tan 3A=\pi \tan 3A\end{array}$