Solve complex equation $\left(\frac{8}{z^3}\right) - i = 0$

Question

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.

Answer 1

$\dfrac8{z^3}=i\implies z^3=\dfrac8i=-8i$.

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 $1$.

Answer 2

Since $i^{3} = -i$

Then $(\frac{2}{z})^{3} + i^{3} = 0 \iff (\frac{2}{z})^{3} - i = 0 \iff z^{3} = -2^{3}i = -8i $

Then the solutions are $z_{1} = 2i, z_{2} = 2e^{i\frac{2\pi }{3}}i, z_{3} = 2e^{i \frac{4\pi }{3}}i$