<s>[INST] Solve complex equation $\left(\frac{8}{z^3}\right) - i = 0$ from $$(a^3 + b^3) = (a+b)(a^2-ab+b^2)$$
I have $${(2/z)}^3 + i^3 =0$$
I have $$\left(\frac{2}{z} + i\right)\left(\left(\frac{2}{z}\right)^2-(2/z)(i)-1)\right) = 0$$
i.e.$ \left(\frac{2}{z}\right)+i = 0 $ or $ \left(\left(\frac{2}{z}\right)^2-(\frac{2}{z})(i)-1\right) = 0$
......
but I'm not sure that is the correct answer;
help me please. 
Thank you.
 [/INST] $\dfrac8{z^3}=i\implies z^3=\dfrac8i=-8i$.
Since $2^3=8$ and $i^3=-i$, it is apparent that $z=2i, 2i\omega$, or $2i\omega^2$, 
where $\omega$ is a primitive cube root of $1$.
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