<s>[INST] Solve the equation $z^3=z+\overline{z}$ I have been trying to solve an equation $z^3=z+\overline{z}$, where $\overline{z}=a-bi$ if $z=a+bi$. But I cant find any clues on how to move forward on that one. Please help.
 [/INST] Note that if $z=a+ib$ then $z+\overline{z}=2a$ and
$$
z^3=2a
$$
The cube roots of $2a$ are
\begin{align}
\sqrt[3]{2a}\cdot\omega^{0}
=
&
\sqrt[3]{2a}
\\
\sqrt[3]{2a}\cdot\omega^{1}
=
&
\sqrt[3]{2a}\big(\frac{1}{2}+i\frac{\sqrt{3}}{2}\big)
\\
\sqrt[3]{2a}\cdot\omega^{2}
=
&
\sqrt[3]{2a}\big(\frac{1}{2}+i\frac{\sqrt{3}}{2}\big)^{2}=\sqrt[3]{2a}\big(\frac{1}{2}-i\frac{\sqrt{3}}{2}\big)
\end{align}
where $\omega=\frac{1}{2}+i\frac{\sqrt[3]{3}}{2}$ is any cubic root of the unit.The possible values ​​of $ a $ and $ b $ are obtained equaling $ z = a + ib$ to the roots of  given above $2a$.
\begin{align}
\sqrt[3]{2a}\cdot\omega^{0}
=
&
\sqrt[3]{2a}
\\
\sqrt[3]{2a}\cdot\omega^{1}
=
&
\sqrt[3]{2a}\big(\frac{1}{2}+i\frac{\sqrt{3}}{2}\big)^{1}
\\
\sqrt[3]{2a}\cdot\omega^{2}
=
&
\sqrt[3]{2a}\big(\frac{1}{2}+i\frac{\sqrt{3}}{2}\big)^{2}=\sqrt[3]{2a}\big(\frac{1}{2}-i\frac{\sqrt{3}}{2}\big)
\end{align}
More explicitly
\begin{align}
a+ib
=
&
\sqrt[3]{2a}
\\
a+ib
=
&
\frac{1}{2}\sqrt[3]{2a}+i\frac{\sqrt{3}}{2}\sqrt[3]{2a}
\\
a+ib
=
&
\frac{1}{2}\sqrt[3]{2a}-i\frac{\sqrt{3}}{2}\sqrt[3]{2a}
\end{align}
</s>