The value of

$\cos ^{3} \frac{\pi}{8}\left[4 \cos ^{3} \frac{\pi}{8}-3 \cos \frac{\pi}{8}\right]$

$+\sin ^{3} \frac{\pi}{8}\left[3 \sin \frac{\pi}{8}-4 \sin ^{3} \frac{\pi}{8}\right]$

$=4 \cos ^{6} \frac{\pi}{8}-4 \sin ^{6} \frac{\pi}{8}-3 \cos ^{4} \frac{\pi}{8}+3 \sin ^{4} \frac{\pi}{8}$

$=4\left[\left(\cos ^{2} \frac{\pi}{8}-\sin ^{2} \frac{\pi}{8}\right)\right]$

$\left[\left(\sin ^{4} \frac{\pi}{8}+\cos ^{4} \frac{\pi}{8}+\sin ^{2} \frac{\pi}{8} \cos ^{2} \frac{\pi}{8}\right)\right]$

$-3\left[\left(\cos ^{2} \frac{\pi}{8}-\sin ^{2} \frac{\pi}{8}\right)\left(\cos ^{2} \frac{\pi}{8}+\sin ^{2} \frac{\pi}{8}\right)\right]$

$=\cos \frac{\pi}{4}\left[4\left(1-\sin ^{2} \frac{\pi}{8} \cos ^{2} \frac{\pi}{8}\right)-3\right]$

$=\frac{1}{\sqrt{2}}\left[1-\frac{1}{2}\right]=\frac{1}{2 \sqrt{2}}$