#z = root[3]{-8} = (-8)^{1/3}#
#z^3 = -8 #
#z^3 + 8 = 0#
It's easy to see #z=-2# is a root so #z+2# is a factor. We get the other factor by dividing:
# \ quad quad quad quad quad quad z^2 -2 z + 4 #
# z+2 | z^3 + 0 z^2 + 0 z + 8 #
# quad quad quad quad quad quad z^3 + 2z^2 #
# quad quad quad quad quad quad quad quad -2z^2 + quad quad quad quad quad 8#
# quad quad quad quad quad quad quad quad -2z^2 - 4z #
# quad quad quad quad quad quad quad quad quad quad quad quad quad quad quad quad quad 4z + 8 #
# quad quad quad quad quad quad quad quad quad quad quad quad quad quad quad quad quad 4z + 8 #
# quad quad quad quad quad quad quad quad quad quad quad quad quad quad quad quad quad quad quad quad quad quad 0 #
#z^3 +8 = (z+2)(z^2 - 2z + 4) = 0#
#z = -2# or #z = 1 \pm sqrt{1-4} = 1\pm i sqrt{3} #
Check:
# (1+i sqrt{3})^3 = (1+ isqrt{3} )(1+i sqrt {3})^2 #
#= (1+ i sqrt{3})(-2 + 2i sqrt{3}) = (-2 - 6) - 2i sqrt{3} + 2i sqrt{3} #
# = -8 quad sqrt#
