<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] This is akin particularly to the methods described by Elias , Sujaan Kunalan and nbubis :
The sum of a complex number and its conjugate produce a pure real number,
$$ z \ + \ \overline{z} \ = \ 2a \ = \ 2a \ \cdot \ cis (0) = \ 2a \ \cdot \ cis (2 \pi)  \ , $$
$ \ cis( \theta ) \ $ being the abbreivated form of $ \ \cos \theta \ + \ i \sin \theta \ \ . $  DeMoivre's Theorem then gives us
$$ z^3 \ = \ r^3  \cdot \ cis(3 \theta) \ = \ 2a \ \cdot \ cis (2 \pi) , $$
with $ \ r \ $ being the modulus of $ \ z \ $ and $ \ \theta \ $ its argument.  We thus obtain  $ \ r^3 = 2a \ $  and the "trigonometric" equation  $ \ cis(3 \theta) \ = \ cis (2 \pi) , $
which is solved (within the "principal circle", $ \ 0 \ \le \ \theta \ < \ 2 \pi \ $ ) by  $ \ \theta \ = \ \frac{0 + 2k \pi}{3} \ , \ \text{for} \ k = 0,1,2 , $
which is described by what is also called DeMoivre's Theorem for roots. 
EDIT:  Oops, need to take this a bit further.
So there are three apparent solutions,
$$z \ = \ (2a)^{1/3} \ , \ (2a)^{1/3} \cdot cis(\frac{2\pi}{3}) \ , \ \text{and} \ (2a)^{1/3} \cdot cis(\frac{4\pi}{3}) \ . $$
For all three, however, we have $ \ z \ + \ \overline{z} \ = \ 2 \cdot (2a)^{1/3} \ , $ which must also equal $ \ 2a \ , $ which is real.   Hence, $ \ (2a)^{1/3} \ = \ a \ \ \Rightarrow \ \ a^{1/3} \ \cdot \ (a^{2/3} - 2^{1/3}) = 0 \ ,  $ which is solved by $ \ a = 0 \ , \ \sqrt{2} \ , \ \text{and} \ -\sqrt{2} \ , $
as found by other posters.  So there are three solutions are pure real numbers.
FURTHER EDIT:  Hmmm, not quite done, on thinking about this a little more.  We've established moduli for solutions, but we should also look at the results for   $ \ (2a)^{1/3} \cdot cis(\frac{2\pi}{3}) \ , \ \text{and} \ (2a)^{1/3} \cdot cis(\frac{4\pi}{3}) \ . $
For $ \ a = \sqrt{2} \ , \ z \ =  \ \sqrt{2} \cdot cis(\frac{2\pi}{3}) \  \ \text{and} \ \sqrt{2} \cdot cis(\frac{4\pi}{3}) \ $, with $ \ \overline{z} \ =  \ \sqrt{2} \cdot cis(\frac{4\pi}{3}) \  \ \text{and} \ \sqrt{2} \cdot cis(\frac{2\pi}{3}) \ , $ respectively, so
$ \ z + \overline{z} \ = \ -2 \sqrt{2} \  , $    but $ \ z^3 \ = \ ( \sqrt{2})^3 \ \cdot \ cis(3 \ \cdot \frac{2 \pi}{3}) \ =  \ 2 \sqrt{2} \ , \ $ and likewise for the argument $ \frac{4 \pi}{3} \ . $
The same thing happens for $ \ a = -\sqrt{2} \ , $  since we have $ \ z \ =  \ -\sqrt{2} \cdot cis(\frac{2\pi}{3}) \ =  \ \sqrt{2} \cdot cis(\frac{5\pi}{3}) \ ,  $ for which $ \ \overline{z} \ =  \ \sqrt{2} \cdot cis(\frac{\pi}{3}) \  $   so,
$ \ z + \overline{z} \ = \ 2 \sqrt{2} \  \   , $ while
$$z^3 \ = \ ( -\sqrt{2})^3 \ \cdot \ cis(3 \ \cdot \frac{2 \pi}{3}) \ = \ (-2 \sqrt{2}) \cdot cis(2 \pi) \ = \ (-2 \sqrt{2}) \cdot (+1) \ = \ -2 \sqrt{2} \ , $$ 
and similarly for   $ \ z \ =  \ -\sqrt{2} \cdot cis(\frac{4\pi}{3}) \ =  \ \sqrt{2} \cdot cis(\frac{\pi}{3}) \ .  $
So it would seem that all of the complex values really are extraneous (as alex.jordan points out).
</s>