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?

Almost there (from your 5th line):
$$
\sin x (4\cos^2 x - 1) = \sin x (3\cos^2 x + \cos^2 x  - 1)
$$
$$
= \sin x (3\cos^2 x + (1-\sin^2 x)  - 1) = 3\sin x\cos^2 x-\sin^3x.
$$
