a0 = ∛8 [ cos [ (pi/2)/3] + i *sin [(pi/2)/3 ] = ∛8 [ cos [ (pi/6)] + i *sin [(pi/6) ] =
2 [ √3/2 + 1/2 * i ] = √3 + 1i
a1 = ∛8 [ cos ( (pi/2)/3 + 2pi/3 ) + i sin ( (pi/2 )/3 + 2pi/3 ) ] =
2 [ cos (5pi/6) + i sin (5pi/6) ] = 2 [ - √3/2 + (1/2)*i ] = - √3 +1i
a2 = ∛8 [ cos ( (pi/2)/3 + 4pi/3 ) + i sin ( (pi/2 )/3 + 4pi/3 ) ] =
2 [ cos(3pi/2) + i sin (3pi/2) ] = 2 [ -1 i ] = -2i
This answer is not mine! the orig solution can be found with a google search :)
