$The value of {\cos }^3\frac{\pi }{8}\cos \frac{3\pi }{8}+{\sin }^3\frac{\pi }{8}\sin \frac{3\pi }{8} is$

$\begin{array}{rcl} & & {\mathrm{co}}^3\frac{\mathrm{\pi }}{8} \cos \frac{3\mathrm{\pi }}{8}+{\sin }^3 \frac{\mathrm{\pi }}{8} \sin \frac{3\mathrm{\pi }}{8}\\ & & \frac{1}{4}\left[4{\cos }^3\frac{\mathrm{\pi }}{8} \cos \frac{3\mathrm{\pi }}{8}+4{\sin }^3\frac{\mathrm{\pi }}{8} \sin \frac{3\mathrm{\pi }}{8}\right]\\ & =& \frac{1}{4}\left[\left(\cos \frac{3\mathrm{\pi }}{8}+3\cos \frac{\mathrm{\pi }}{8}\right) \cos \frac{3\mathrm{\pi }}{8}+\left(3\sin \frac{\mathrm{\pi }}{8}-\sin \frac{3\mathrm{\pi }}{8}\right) \sin \frac{3\mathrm{\pi }}{8}\right]\\ & =& \frac{1}{4}\left({\cos }^2\frac{3\mathrm{\pi }}{8}+3\cos \frac{3\mathrm{\pi }}{8} \cos \frac{\mathrm{\pi }}{8}+3\sin \frac{3\mathrm{\pi }}{8} \sin \frac{\mathrm{\pi }}{8}-{\sin }^2\frac{3\mathrm{\pi }}{8}\right)\\ & & \frac{1}{4}\left[\cos \frac{6\mathrm{\pi }}{8}+3\cos \frac{2\mathrm{\pi }}{8}\right]\\ & =& \frac{1}{4}\left[-\cos \frac{2\mathrm{\pi }}{8}\times 3\cos \frac{2\mathrm{\pi }}{8}\right]\\ & =& \frac{1}{4}\left(2\cos \frac{\mathrm{\pi }}{4}\right)=\frac{1}{2}·\frac{1}{\sqrt{2}}\\ & =& \frac{1}{2\sqrt{2}}\end{array}$