\[\begin{array}{l} \sin \frac{\pi }{{16}}{\cos ^3}\frac{\pi }{{16}} - {\sin ^3}\frac{\pi }{{16}}\cos \frac{\pi }{{16}} = \sin \frac{\pi }{{16}}\cos \frac{\pi }{{16}}\left( {{{\cos }^2}\frac{\pi }{{16}} - {{\sin }^2}\frac{\pi }{{16}}} \right) = \hfill \\ \frac{1}{2}\sin \frac{\pi }{8}\cos \frac{\pi }{8} = \frac{1}{4}\sin \frac{\pi }{4} = \frac{1}{4} \cdot \frac{{\sqrt 2 }}{2} = \frac{{\sqrt 2 }}{8}. \hfill \\ \end{array}\]

\[\frac{{\sqrt 2 }}{8}\]