Answer
$1$, $-\frac{1}{2}+\frac{\sqrt 3}{2}i$, $-\frac{1}{2}-\frac{\sqrt 3}{2}i$
Work Step by Step
Step 1. Let $z=1$; we have $z=cos0+i\ sin0$
Step 2. Based on De Moivre's Theorem, we have the cube roots as
$z_k=\sqrt[3] 1(cos\frac{2k\pi+0}{3}+i\ sin\frac{2k\pi+0}{3})$
where $k=0,1,2$
Step 3. For $k=0$, we have $z_0=1$
Step 4. For $k=1$, we have
$z_1=cos\frac{2\pi}{3}+i\ sin\frac{2\pi}{3}=-\frac{1}{2}+\frac{\sqrt 3}{2}i$
Step 5. For $k=2$, we have
$z_2=cos\frac{4\pi}{3}+i\ sin\frac{4\pi}{3}=-\frac{1}{2}-\frac{\sqrt 3}{2}i$
