Find the cube roots of 8(cos 216° + i sin 216°).

Solution:

Let Z = 8(cos 216° + i sin 216°)

Taking cube root on both sides

Z^1/3 = (8)^1/3 [cos216° + i sin 216° ]^1/3

Using De- Moivers theorem 

If z = r [cos θ + i sinθ] then zⁿ = rⁿ [cos nθ + i sin nθ]

z^1/3 = (2³)^1/3 [cos(216°/3) + i sin (216°/3)]

z^1/3 = 2 [ cos72°+ i sin72°]

Find the cube roots of 8(cos 216° + i sin 216°).

Summary:

The cube root of 8(cos 216° + i sin 216°) is 2 [ cos72°+ i sin72°].