$\sin 3\mathrm{\theta }-{\cos }^3\mathrm{\theta }+\cos 30\cdot {\sin }^3\mathrm{\theta }=$

$$\begin{gathered} =\left(3\sin \mathrm{\theta }-4{\sin }^3\mathrm{\theta }\right){\cos }^3\mathrm{\theta }+\left(4{\cos }^3\mathrm{\theta }-3\cos \mathrm{\theta }\right){\sin }^2\mathrm{\theta }=3\sin {\mathrm{\theta cos}}^2\mathrm{\theta }-3\cos \mathrm{\theta sin}\mathrm{\theta } \\ =3\sin \mathrm{\theta cos}\mathrm{\theta }\left({\cos }^2\mathrm{\theta }-{\sin }^2\mathrm{\theta }\right)=\frac{3}{2}\cdot \sin 2\mathrm{\theta }\cdot \cos 2\mathrm{\theta }=\frac{3}{4}\sin 4\mathrm{\theta } \end{gathered}$$