How does $x^3 - \sin^3 x$ become $x^3 + \frac{1}{4}\sin{3x}-\frac{3}{4}\sin x$? I was going through answers on this question and came across this answer and I was wondering how the user arrived at the first line where they state:
$$f(x) \equiv x^3 - \sin^3 x = x^3 + {1 \over 4} \,\sin {3x} - {3 \over 4}\,\sin x$$

How does $x^3 - \sin^3 x$ become $x^3 + \frac{1}{4}\sin{3x}-\frac{3}{4}\sin x$?

Are they using some simple identity or is there some other observation happening?
Thanks!
HINT: $$\sin{3x}=3\sin {x} -4\sin^3 x \\ \sin {(A+B)}=\sin A \cos B + \cos A \sin B$$
Hint:
$$\sin x =\frac{opposite}{hypotenuse}$$