Find All Complex solutions for $z^3+3i\bar{z}=0$ Find All Complex solutions for $z^3+3i \bar z =0$.
I tried substituting $z=a+bi$ and simplifying as much as I can
and this is what I ended up with: $a^3-2b^2a+3b+i(3ba^2-b^3+3a)=0$.
I just did not understand how do I get the values of $z$ from this equation

This is a commentary on all of the existing answers, rather than a suggested method of solution. It seems fashionable to state that there are $9$ solutions, when in reality there are $5$ solutions $z$ to the equation $$z^3-3i\overline z = 0$$
Writing $z=x+iy$, we obtain the following polynomial equations by taking real and imaginary parts of the given equation:
\begin{align*}
&x^3-3xy^2+3y = 0 \\
&3x^2y+3x-y^3 = 0
\end{align*}
By Bézout's Theorem, the corresponding projective curves have $9$ intersections, counting multiplicity. However, we are seeking real solutions $(x,y)$ in the affine plane, so there could be fewer than $9$ solutions to the equation $z^3-3i\overline z = 0$.
By standard polynomial algebra, this system is equivalent to
\begin{align*}
&9x-8y^7+51y^3 = 0 \\
&64y^9-432y^5+81y=0
\end{align*}
Factoring $y$ out of the second equation and performing the substitution $Y=y^4$, we obtain
\begin{equation*}
64Y^2-432Y+81
\end{equation*}
The roots are $$Y = \frac{27}8 \pm \frac94\sqrt2$$
These are positive real numbers, and so the solutions for $y$ are
\begin{equation*}
y = 0, \pm\left(\frac{27}8 \pm \frac94\sqrt2\right)^{1/4}
\end{equation*}
For each $y$, the coordinate $x$ is uniquely determined and real. This means that there are $5$ solutions to the equation,
$$z^3-3i\overline z = 0$$
