I do know how to calculate the cubed root of a complex number....like if I'm given that $x^3=p$, where $p$ is a complex number, then $$x= r^{1/3}\left(\cos\left(\frac{2k\pi+m}{3}\right) + i\sin \left(\frac{2k\pi+m}{3}\right)\right)$$ where $p$ is $r\left(\cos m +i\sin m\right)$ and $k=0,1,2$
But can I write it in this way?? $$x^3=p \implies x=p^{1/3},\,p^{1/3}w , \,p^{1/3} w^2, \dots$$ where $w$ is the cubed root of unity? I thought that if I write the roots in this way and put any value of $p^{1/3}$, I get all the three values of $x$ what I got using the first method...thanks
HINT: use that $$x^3-a=(x-a^{1/3})(x^2+xa^{1/3}+a^{2/3})$$