<s>[INST] How to find all values of $z$ such that $z^3=−8$ The question is
Solve the equation
$z^3=-8$
My attempt
I attempt to write it out in polar co-ordinates
since $z = r(\sin(x) + i\sin(x)) \\ z^3 = r^3(\cos(3x) + i\sin(3x))$
then
$r^3\sin(3x) = -8$ and $r^3\cos(3x) = 0$
but from here I'm not really sure where to go , I've searched up the solution to this before and people have written $r^3 = 8$ so $\cos(3x) = -1$ and $\sin(3x) = 0$ .
 [/INST] We could also apply the "triple-angle" identities to write $ \ \sin(3 \theta) \ = \ \sin \theta · (3 - 4 \sin^2 \theta) \ = \ 0 \ \  $ and $ \ \cos(3 \theta) \ = \ 4 \cos^3 \theta - 3 \cos \theta \ = \ -1 \ \  . $
From the first of these, we have either $ \ \sin \theta \ = \ 0 \ \Rightarrow \ \theta \ = \ 0 \ , \ \pi \ \ $ or $  \ \sin^2 \theta \ = \ \frac34 \  \Rightarrow \ \cos^2 \theta \ = \ \frac14 \ \ . $  Inserting these results into the second equation, we find that $ \ \theta \ = \ 0  \ $ is not a solution, but $ \ \theta \ =  \ \pi \ $ is $ \ ( \ \cos(3 \pi) \ = \ 4 · [-1]^3  - 3 ·[-1] \ = \ -1 \   ) \ . $  For  $ \ \cos  \theta \ = \ \pm \frac12 \ \ , $ we see that $ \ 4 · \left[-\frac12 \right]^3  - 3 ·\left[-\frac12 \right] \ = \ +1 \    \ , $ but $ \ 4 · \left[+\frac12 \right]^3  - 3 ·\left[+\frac12 \right] \ = \ -1 \    \ . $
Hence the three angle solutions are $ \ \theta \ = \ \pi \ , \ \pm \ \frac{\pi}{3} \ \ , $ which gives the three solutions to the original equation (the three complex cube-roots of $ \ -8 \ $ as
$$ 2·(cis \ \pi) \ \ = \ \ -2 \ \ \ , \ \  \  2·\left(cis \ \pm \frac{\pi}{3} \right) \ \ = \ \ 1 \ \pm \ i\sqrt3 \ \ . $$
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