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?

Also it can be proved using complex numbers.
$(\cos x+i\sin x)^3=\cos 3x+i\sin 3x $ (Moivre's formula)
$(\cos x+i\sin x)^3=(\cos^3x-3\cos x\sin^2x)+i(3\cos^2 x\sin x-\sin^3x)$ (binom)
Comparing these two formulas, we obtain 
$\cos 3x=\cos^3x-3\cos x\sin^2x=\cos^3x-3\cos x(1-\cos^2x)=\ldots$
$\sin 3x =3\cos^2 x\sin x-\sin^3x=\ldots$
