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If $x, y, z$ are complex numbers, how can I solve this system of equations \begin{cases} x(x-y)(x-z)=3;\ \\y(y-z)(y-x)=3;\ \\z(z-x)(z-y)=3. \end{cases}

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I guess you should divide first equation by the second one,the second one by the third one...etc – pedja Dec 1 '11 at 9:05
I have tried, however I can not get conclusion. – Chen Dec 1 '11 at 10:39
,I don't see any other approach to this problem... – pedja Dec 1 '11 at 10:50
The answer consists of the roots of the equation $z^3=1$. – Raskolnikov Dec 1 '11 at 11:24

I suggest following approach, take the first equation and the second and multiply the right hand side of one with the left hand side of the other and visa versa to get after simplyfing:

$$3x(x-z)=-3y(y-z) \; .$$

Rearranging, you can write this as

$$x^2+y^2=z(x+y) \; .$$

Doing this to the other pairs of equations you get the new system:

$$\begin{cases} x^2+y^2=z(x+y) \; ,\ \\y^2+z^2=x(y+z) \; ,\ \\x^2+z^2=y(x+z) \; . \end{cases}$$

Substracting the first of these with the third, you get

$$z^2-y^2=x(y-z) \; ,$$

and adding that to the second equation gives

$$z^2=xy \; .$$

So you'll end up with the system

$$\begin{cases} x^2=yz \; ,\ \\y^2=xz \; ,\ \\z^2=xy \; . \end{cases}$$

multiplying these equations by the appropriate factor, you can see that

$$x^3=y^3=z^3=xyz \; .$$

This is satisfied by the roots of a polynomial equation of the form $z^3=z_0$.

Let's call one of these roots $r$, then the other roots are $re^{2\pi i/3}$ and $re^{4\pi i/3}$. Filling this in the first of your original equations will determine that $r$ should be a cube root of unity. Therefore the solutions will be: $x=1, y=e^{2\pi i/3}$ and $z=e^{4\pi i/3}$ and any permutation of these.

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