If $\mathrm{f}\left(\mathrm{\theta }\right)={\sin }^3\mathrm{\theta }+{\sin }^3\left(\mathrm{\theta }+\frac{2\mathrm{\pi }}{3}\right)+{\sin }^3\left(\mathrm{\theta }+\frac{4\mathrm{\pi }}{3}\right)$ then the value of $\mathrm{f}\left(\frac{\mathrm{\pi }}{18}\right)+\mathrm{f}\left(\frac{7\mathrm{\pi }}{18}\right)$ is

$\begin{array}{r}\sin \mathrm{\theta }+\sin \left(\mathrm{\theta }+\frac{2\mathrm{\pi }}{3}\right)+\sin \left(\mathrm{\theta }+\frac{4\mathrm{\pi }}{3}\right)=0\\ \Rightarrow {\sin }^3\mathrm{\theta }+{\sin }^3\left(\mathrm{\theta }+\frac{2\mathrm{\pi }}{3}\right)+{\sin }^3\left(\mathrm{\theta }+\frac{4\mathrm{\pi }}{3}\right)\end{array}$

$\begin{array}{l}=3\left(\sin \mathrm{\theta }\cdot \sin \left(\mathrm{\theta }+\frac{2\mathrm{\pi }}{3}\right)\cdot \sin \left(\mathrm{\theta }+\frac{4\mathrm{\pi }}{3}\right)\right)\\ \mathrm{f}\left(\mathrm{\theta }\right)=\frac{-3}{4}\sin 3\mathrm{\theta }\\ \therefore \mathrm{f}\left(\frac{\mathrm{\pi }}{18}\right)+\mathrm{f}\left(\frac{7\mathrm{\pi }}{18}\right)=\frac{-3}{4}\left[\sin \frac{\mathrm{\pi }}{6}+\sin \frac{7\mathrm{\pi }}{6}\right]=0\end{array}$