an only slighter easier way:
suppose we already know how to find the 3 cube roots of 1 (by, say, using trigonometry), call these: 1, ω, and ω
^{2} (=
).
(it should be easy to see that if ω
^{3} = 1, and ω ≠ 1, then (ω
^{2})
^{3} = (ω
^{3})
^{2} = (1)(1) = 1, as well).
(we can find these by factoring x
^{3} - 1 = (x - 1)(x
^{2} + x + 1), also).
then (x + 2)(x + 2ω)(x + 2ω
^{2}) = (x + 2)(x
^{2} + (2ω + 2ω
^{2})x + 4ω
^{3})
= (x + 2)(x
^{2} + 2(ω + ω
^{2})x + 4), and we know from above that: ω + ω
^{2} = -1, so:
= (x + 2)(x
^{2} - 2x
^{2} + 4) = x
^{3} + 8
so the three cube roots of -8 are -2,-2ω, and -2ω
^{2}.