z^3=\Big (\frac{\sqrt{3}}{2}i-\frac{1}{2}\Big)^3=\Big(\frac{\sqrt{3}}{2}i\Big)^3-3\Big(\frac{\sqrt{3}}{2}i\Big)^2\cdot{\frac{1}{2}}+3\frac{\sqrt{3}}{2}i\cdot{\Big (\frac{1}{2}\Big )^2}-\Big (\frac{1}{2}\Big)^3=-\frac{3\sqrt{3}}{8}i+\frac{9}{8}+\frac{3\sqrt{3}}{8}i-\frac{1}{8}=1