Question

If $x+y+z=xyz$, prove that $\frac{3x-x^3}{1-3x^2}+\frac{3y-y^3}{1-3y^2}+\frac{3z-z^3}{1-3z^2}=\frac{3x-x^3}{1-3x^2}\cdot \frac{3y-y^3}{1-3y^2}\cdot \frac{3z-z^3}{1-3z^2}$.

Answer 1

$\tan (3A+3B+3C)=0$
$\therefore \frac{S_1-S_3}{1-S_2}=0$ so that $S_1=S_3$
or $\tan 3A+\tan 3B+\tan 3C=\tan 3A\cdot \tan 3B\cdot 3C$
Now put $\tan 3A=\frac{3\tan A-{\tan }^{3}A}{1-3{\tan }^{2}A}=\frac{3x-x^3}{1-3x^2}$ etc.

SImilarly we can write for $\tan 3B$ and $\tan 3C$.

$\therefore \frac{3x-x^3}{1-3x^2}+\frac{3y-y^3}{1-3y^2}+\frac{3z-z^3}{1-3z^2}=\frac{3x-x^3}{1-3x^2}\cdot \frac{3y-y^3}{1-3y^2}\cdot \frac{3z-z^3}{1-3z^2}$.