8(cos 264º + j sin 264º) = 8 cis(264 + 360*n), n=0, +/-1, +/-2, ..

So:

(8(cos 264º + j sin 264º))^{1/3} = 2 cis(264/3 + 360*n/3) n =...

............= 2 cis(88 + 120*n) n=0, +/-1, +/-2, ..

Now only three of these are distinct, and taking n=0, -1, +1 will give us

a complete set.

These are:

2 cis(88), 2 cis(-32), 2 cis(208).

The principal root is the one with positive imaginary part, which is 8 cis(88).

RonL