<s>[INST] Show that $\sin(3x)$ is equivalent to $3\sin(x)\cos^2(x)-\sin^3(x)$ 
The expression $\sin(3x)$ is equivalent to:
A. ...

My book states the right answer is B which is $3\sin(x)\cos^2(x)-\sin^3(x)$.
I tried:
$$\sin(x)\cos(2x)+\cos(x)\sin(2x) = \\
\sin(x)(2\cos^2(x)-1)+\cos(x)\cdot2\sin(x)\cos(x) = \\
2\cdot\sin(x)\cos(x)\cos(x)-\sin(x)+\cos(x)2\sin(x)\cos(x) = \\
\sin(x)(2\cos^2(x)-1+2\cos^2(x)) = \\
\sin(x)(4\cos^2(x)-1) = \\
4\cos^2(x)\sin(x)-\sin(x) = ???$$
What do I do next?
 [/INST] Notice that \begin{equation}\begin{split}\sin(3x)  &= 4\cos^2 x \sin x - \sin x\\&= 3\sin x \cos^2 x +(\sin x \cos^2x -\sin x )\\&=3\sin x \cos^2 x + \sin x\cdot( \cos^2x - 1)\\&= 3\sin x \cos^2 x -\sin^3 x \end{split}\end{equation}
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