Verifying $\sin 4θ=4\cos^3 θ \sin θ - 4\cos θ \sin^3θ$ $$\sin 4θ=4\cos^3 θ \sin θ - 4\cos θ \sin^3θ.$$

Ηere is what I have so far
$$\sin 4θ = 2\sin 2θ \cos 2θ = 4\sin θ \cos θ \cos 2θ.$$

Not sure if this is the correct path I should take to solve this problem. I have been stuck hard for about an hour now.
$$4\cos^3\theta\sin\theta-4\sin^3\theta\cos\theta=4\sin\theta\cos\theta(\cos^2\theta-\sin^2\theta)=2\sin2\theta\cos2\theta=\sin4\theta.$$
$$e^{ix} = i\sin{x}+\cos{x}$$
$$e^{4ix} = (i\sin{x}+\cos{x})^4$$
$$i\sin{4x}+\cos{4x} = \sin^4{x}-4i\sin^3x\cos{x}-6\sin^2x\cos^2{x}+4i\sin x\cos^3{x} + \cos^4x$$
As all coefficients are real, we can say the imaginary parts of each side are equal:
$$i\sin{4x} = -4i\sin^3x\cos{x}+4i\sin x\cos^3{x}$$
$$\sin{4x} = 4\sin x\cos^3{x}-4\sin^3x\cos{x}$$