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Consider system of equations $x(x-y)(x-z)=3$, $y(y-x)(y-z)=3$, $z(z-y)=3$ where $x,y,z\in\mathbb C$. Then which of the following is/are True?

A. There are different solutions.

B. sum $(x+y+z)$ in any solution is Zero.

C. No two of $x,y,z$ can be simultaneously real

D. Any solution lies on a straight line.

I tried various ways but I couldn't figure it out. According to Wolfram Alpha, there are $7$ different solutions to it.

Sam
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  • Oof, if I put the problem through a Groebner basis calculation, it gives a degree 7 polynomial in $z$ (and it does appear that polynomial has nonzero discriminant), and then it can express $x$ and $y$ in terms of $z$. – Daniel Schepler Oct 06 '23 at 19:10
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    On the other hand, if I change the last equation to $z(z-x)(z-y)=3$ (for symmetry with the others), then I get a simple Groebner basis of $x+y+z, y^2+yz+z^2, z^3-1$. – Daniel Schepler Oct 06 '23 at 19:13
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    To follow up on @DanielSchepler's observation: the solutions of the symmetric system are the six permutations of $(x,y,z)=(1,e^{2\pi i/3},e^{4\pi i/3})$. This in particular would mean that $x+y+z=0$. – Semiclassical Oct 06 '23 at 19:40

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