Find the cube roots of 125(cos 288° + i sin 288°).

Solution:

Let Z = 125(cos 288° + i sin 288°)

Taking cube root on both sides

Z^1/3 = (125)^1/3 [cos288° + i sin 288° ]^1/3

Using De- Moivers theorem 

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

z^1/3 = (5³)^1/3 [cos(288°/3)+ i sin (288°/3)]

z^1/3 = 5 [ cos 96°+ i sin 96°]

Find the cube roots of 125(cos 288° + i sin 288°).

Summary: 

The cube roots of 125(cos 288° + i sin 288°) is 5 [ cos 96°+ i sin 96°]