Solving complex equation ${{\left( z+2\bar{z} \right)}^{3}}=1$ Hey i am stuck tring to solve :
$${{\left( z+2\bar{z} \right)}^{3}}=1$$
I used Binomial theorem to expand the equation to :
$${{z}^{3}}+6{{z}^{2}}\bar{z}+12z{{\bar{z}}^{2}}+8{{\bar{z}}^{3}}=1$$
and i am not sure how should i continue

Here $$(z+2\bar{z})^3 = 1\Rightarrow (z+2\bar{z}) = (1)^{\frac{1}{3}} = 1,\omega,\omega^2.$$
Where $\omega,\omega^2$ are cube root of unity.
So $$z+2\bar{z} = 1,\omega,\omega^2.$$
Now Put $z=x+iy$ and $\bar{z} = x-iy$
$\bullet\; $ If $z+2\bar{z} = 1\;,$ Then $x+iy+2(x-iy) = 1\Rightarrow 3x-iy=1+0\cdot i$
So we get $\displaystyle x = \frac{1}{3}$ and $y=0$
$\bullet\; $ If $z+2\bar{z} = \omega\;,$ Then $\displaystyle x+iy+2(x-iy) = -\frac{1}{2}+i\frac{\sqrt{3}}{2}\Rightarrow 3x-iy=-\frac{1}{2}+i\frac{\sqrt{3}}{2}$
So we get $\displaystyle x = -\frac{1}{6}$ and $\displaystyle y=-\frac{\sqrt{3}}{2}$
$\bullet\; $ If $z+2\bar{z} = \omega\;,$ Then $\displaystyle x+iy+2(x-iy) = -\frac{1}{2}-i\frac{\sqrt{3}}{2}\Rightarrow 3x-iy=-\frac{1}{2}-i\frac{\sqrt{3}}{2}$
So we get $\displaystyle x = -\frac{1}{6}$ and $\displaystyle y=+\frac{\sqrt{3}}{2}$
So we get $$\displaystyle z = \left\{\frac{1}{3}+0\cdot i\;\;,-\frac{1}{6}-\frac{\sqrt{3}}{2}\cdot i\;\;,-\frac{1}{6}+\frac{\sqrt{3}}{2}\cdot i\right\}$$
