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.

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