If $\cos^3\frac{\pi}{9}+ \sin^3\frac{\pi}{18} = \dfrac{m}{4} \left( \cos\frac{\pi}{9}+ \sin\frac{\pi}{18}\right)$.Find $m$

$\cos ^{3} \frac{\pi}{9}+\sin ^{3} \frac{\pi}{18}=\frac{m}{4}\left(\cos \frac{\pi}{9}+\sin \frac{\pi}{19}\right)$

$\begin{array}{l}\cos 3 A=4 \cos ^{3} A-3 \cos A \\ \cos ^{3} A=\frac{\cos 3 A+3 \cos A}{4} \\ \sin ^{3} A=\frac{3 \sin A-\sin 3 A}{4}\end{array}$

$\begin{array}{c}\cos ^{3} \frac{\pi}{9}+\sin^{3} \frac{ \pi}{18}=\left[\cos{3} \frac{ \pi}{9}-\sin{3} \frac{ \pi}{18}\right] \frac{1}{4} +\frac{3}{4}\left[\cos \frac{\pi}{9}+\sin \frac{\pi}{18}\right]\end{array}$

$=\left[\cos \frac{\pi}{3}-\sin \frac{\pi}{6}\right] \frac{1}{4}+\frac{3}{4}\left[\cos \frac{\pi}{9}+\sin \frac{\pi}{18}\right]$

$=[\frac{1}{2}-\frac{1}{2}]^{\frac{1}{4}}+\frac{3}{4}\left[\cos \frac{\pi}{9}+\sin \frac{\pi}{18}\right]$

$\cos ^{3} \frac{\pi}{9}+\sin ^{3} \frac{\pi}{18}=\frac{3}{4}\left[\cos \frac{\pi}{9}+\sin \frac{\pi}{18}\right]$