<s>[INST] solve complex equation for $x$ and $y$ How to solve this?
I have tried to put all $z$s on one side, 
but I don't have an idea to continue.
$$z^3-i(z-2i)^3=0$$
 [/INST] $$
z^3 = i\cdot(z-2i)^3.
$$
Therefore $z$ is one of the three cube roots of $i(z-2i)^3$.
Observe that $(-i)^3 = i$ so $z$ is one of the three cube roots of $(-i)^3 (z - 2i)^3 = (-2-iz)^3$.
One of those three cube roots is $-2-iz$.
So either $z=-2-iz$ or $z$ is one of the other two cube roots.
If $z=-2-iz$ then $z(1+i) = -2$ so $z=-2/(1+i) = i-1$.
Since $z$ is one of the three cube roots of its cube and it's one of the three cube roots of $-2-iz$, the three cube roots of $i-1$ are what we seek.
What are the three cube roots of $i-1 = \sqrt{2}\left(\cos(135^\circ) + i\sin(135^\circ)\right)$.  Here we invoke de Moivre's theorem and seek the three numbers that are $135^\circ/3$ and this division is done modulo $360^\circ$.  That gives us
$$
45^\circ \text{ or } {45}^\circ \pm \frac{360^\circ} 3 = 45^\circ \text{ or } {-75}^\circ \text{ or } {+165}^\circ.
$$
Modulo $360^\circ$, that is the same as
$$
45^\circ \text{ or } 165^\circ \text{ or } 285^\circ,
$$
all differing from each other by a third of a circle.  So we have:
\begin{align}
\sqrt 2\left(\cos45^\circ + i \sin45^\circ\right) & = 1+i \\[10pt]
\sqrt2 \left(\cos165^\circ + i\sin165^\circ\right) & = \cdots \\[10pt]
\sqrt 2\left(\cos 185^\circ + i \sin185^\circ\right) & = \cdots
\end{align}
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