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}.