<s>[INST] If $x+y+z=xyz$, find $\frac{3x-x^3}{1-3x^2}+\frac{3y-y^3}{1-3y^2}+\frac{3z-z^3}{1-3z^2}$ I found this question in a maths worksheet of trigonometry (kinda odd, right?), but I dont know how to figure it out.
If $\displaystyle x+y+z=xyz$, find $\displaystyle\frac{3x-x^3}{1-3x^2}+\frac{3y-y^3}{1-3y^2}+\frac{3z-z^3}{1-3z^2}$
First I thought of taking x,y and z as $\displaystyle \tan A, \tan B,$ and $\tan C,$ making $A+B+C=\pi$, but couldnt solve ahead. Any solution not involving trigonometry would do as well.
I also think that this question does not even relate to trigo.....or does it?
 [/INST] We can not make $A+B+C=\pi$
as $\displaystyle \tan(A+B+C)=\frac{\sum\tan A-\tan A\tan B\tan C}{1-\sum \tan A\tan B}$
$\displaystyle\sum\tan A=\tan A\tan B\tan C\implies \tan(A+B+C)=0\implies A+B+C=n\pi$ where $n$ is any integer [Clearly, you have taken a special value$(1)$ of $n$]
Now for any integer $m,$ $\displaystyle m(A+B+C)=mn\pi\implies \tan(mA+mB+mC)=\tan mn\pi=0$
$\displaystyle\implies\sum\tan mA=\tan mA\tan mB\tan mC$
Now set $m=3$ and apply $\tan3x$ formula
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