Is $\sin^3 x=\frac{3}{4}\sin x - \frac{1}{4}\sin 3x$? $$\sin^3 x=\frac{3}{4}\sin x - \frac{1}{4}\sin 3x$$
Is there any formula that tells this or why is it like that?

\begin{align}
\sin(3x)           &= \sin(x+2x)                                                  \tag{1} \\ 
\sin(\alpha+\beta) &= \sin \alpha \cdot \cos \beta + \cos \alpha \cdot \sin \beta \tag{2} \\ 
\sin 2\alpha       &= 2\cdot \sin \alpha \cdot \cos \alpha                        \tag{3} \\ 
\cos 2\alpha       &= \cos^2 \alpha - \sin^2 \alpha                               \tag{4} \\ 
1                  &= \sin^2 \alpha + \cos^2 \alpha                               \tag{5} %% 
\end{align}
If you apply all these formulas you should get:
$$ \sin(3x)=3\cdot \sin x -4\cdot \sin^3 x $$
