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So I was bored, so I decided to look back at my questions and noticed this question of mine asking about trinomial expansion which I thought "hey, what if I created a piecewise equation using trinomial expansion", which made me come up with this:$$ \color{white}{.=} \begin{cases} (x+y+z)^3 & =18 \\ x+y & =z-xy \\ y+z & =xy+y \end{cases} $$which I thought that I might be able to solve. Here is my attempt at solving the piecewise equation:

Since I know that$$x+y=z-xy$$and$$y+z=xy+y$$$$\therefore z=xy$$$$\therefore x+y=xy-xy$$$$\therefore x+y=0$$$$\implies x=-y,y=-x$$Also, since we know that $z^3=18,z=\sqrt[3]{18}$, therefore$$(x+y+\sqrt[3]{18})^3=18$$therefore$$(0+\sqrt[3]{18})^3=18$$and therefore$$x=y=0,z=\sqrt[3]{18}$$However, this is incorrect. Why? Because $0\neq\sqrt[3]{18}$ and $\sqrt[3]{18}\neq0+0$

My second attempt

$$x+\require{cancel}\cancel{y}-\cancel{y}-z=z-xy-xy-y$$$$x-z=z-2xy-y$$$$x-\cancel{z}+\cancel{z}=2z-2xy-y$$$$\therefore x=2z-2xy-y$$$$y=x-2z+2xy$$$$z=x-z+2xy-y$$Then plugging this into $x+y=z-xy$, we get$$x+y=x-z+2xy-y-8xyz+4x^2y^2+2x^2-2xy^2+2y^2-2xz-xy+4z^2$$However, this was beginning to get extremely complicated for me, so I decided to plug it into Wolfram Alpha which gave me these results (all complex)

My question

What could I do to attain the correct solution, since it seems I was unable to do so myself?

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    Firstly you should get three solutions to the equation $z^3=18$, not just one. The other two are complex. Secondly $x+y=0$ does not imply $x=y=0$. When you have your $z$ values, use $z=xy\implies z=-x^2$ to find $x$ and $y$ – David Quinn May 22 '23 at 16:04
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    From the third equation, $ z = xy $. Substituting this into the second equation, $x + y = 0$. Substituting this into the first equation $z^3 = 18$. This is straight forward to solve. – Hosam Hajeer May 22 '23 at 17:28

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