trigonometry - Euler formula and $sin^3$

Using the formula:

$$e^{i\omega t} = \cos {\omega t} + i\sin{\omega t}$$

I would like to prove that:

$$\sin^3\;x = -\frac{\sin{3x} - 3\sin{x}}{4}$$

However I haven't found any approach to this question. Just converting the first formula to $\sin^3$ doesn't seem to help as I'm still getting $\cos^3$ on the other side. Can anyone help me to guide me on the right way?