Question: > 
> following Diophantine equation 
>  $$xy^2+yz^2+zx^2=x^2y+y^2z+z^2x+x+y+z$$
>  ie:$(x-y)(y-z)(z-x)=x+y+z$
> 
> 
> 

where $x,y,z$ are integers.

can find $x+y+z$ 

I tried some values and got some near equalities when $x,y,z$ at least two are equal, $x+y+z=0$,for other case, who have an idea to discuss...plz. Thanks in advance.

Answer: 
Hint:
$$x\sqrt{\frac{a^2-x^2}{a^2+x^2}}=x\sqrt{\frac{1-(\frac{x}{a})^2}{1+(\frac{x}{a})^2}}$$
and use the trigonometric identity
$$\sqrt{\frac{1-\cos \theta}{1+\cos \theta}}=|\tan\left(\frac{\theta}{2}\right)|$$
**Hint:** $~\dfrac{1+\cos2t}2=\cos^2t~$ and $~\dfrac{1-\cos2t}2=\sin^2t.$