Since you're mentioning "all" the cube roots, I'm assuming you're asking about complex roots as well.
In that case, it is more convenient to use De Moivre's form:
[tex]z=8=8(1+0i)=8[\cos(2\pi)+i\sin(2\pi)][/tex]
Which implies
[tex]\sqrt[3]{z}=2\left[\cos\left(\dfrac{2n\pi}{3}\right)+i\sin\left(\dfrac{2n\pi}{3}\right)\right],\quad n=0, 1, 2[/tex]
Which yields the roots
[tex]z_0 = 2\left[\cos\left(0\right)+i\sin\left(0\right)\right] = 2[/tex]
[tex]z_1=2\left[\cos\left(\dfrac{2\pi}{3}\right)+i\sin\left(\dfrac{2\pi}{3}\right)\right]=-1+i\sqrt{3}[/tex]
[tex]z_2=2\left[\cos\left(\dfrac{4\pi}{3}\right)+i\sin\left(\dfrac{4\pi}{3}\right)\right]=-1-i\sqrt{3}[/tex]
