<s>[INST] Expanding $4\sin^3(x)$ using the Complex Exponential How would I go about proving:
$$
4 \sin^3(x)=3\sin(x)-\sin(3x)
$$
Using the complex exponential, i.e.
$$
e^{ix}=\cos(x)+i\sin(x)
$$
 [/INST] $$(\cos x+i\ \sin x)^n=\cos nx+i\ \sin nx....(1)$$
Since we want $\sin^3x$  we  will expand $(\cos x+i\ \sin x)^3$
$(\cos x+i\ \sin x)^3=\cos^3x-3\cos x\sin^2x+i\ (3\cos^2x\sin c-\sin^3x)$
From $(1)$ we have 
$$\cos3x+i\ \sin3x=\cos^3x-3\cos x\sin^2x+i(3\cos^2x\sin x-\sin^3 x)$$
Now equate the imaginary parts and we get
$$\sin3x=3\cos^2x\sin x-\sin^3x$$
$$\sin3x=3(1-\sin^2x)\sin x-\sin^3x$$
$$\sin3x=3\sin x-3\sin^2x\sin x-\sin^3 x$$
$$\sin3x=3\sin x-4\sin^3x$$
Therefore, $$4\sin^3x=3\sin x-\sin3x$$
</s>