${\tan }^6\frac{\pi }{9}-33 {\tan }^4\frac{\pi }{9}+27 {\tan }^2\frac{\pi }{9} =$

$$\begin{gathered} \tan 3\mathrm{A}=\frac{3\tan \mathrm{A}-{\tan }^3\mathrm{A}}{1-3{\tan }^2\mathrm{A}} \\ \tan \frac{\mathrm{\pi }}{3}=\frac{3\tan {\displaystyle \frac{\mathrm{\pi }}{9}}-{\tan }^3{\displaystyle \frac{\mathrm{\pi }}{9}}}{1-3{\tan }^3{\displaystyle \frac{\mathrm{\pi }}{9}}} \\ {\left[\sqrt{3}\left(1-3{\tan }^2\frac{\mathrm{\pi }}{9}\right)\right]}^2={\left[\left(3\tan \frac{\mathrm{\pi }}{9}-\tan \frac{\mathrm{\pi }}{9}\right)\right]}^2 \\ 3\left(1+9{\tan }^4\frac{\mathrm{\pi }}{9}-6{\tan }^2\frac{\mathrm{\pi }}{9}\right)=9{\tan }^2\frac{\mathrm{\pi }}{9}+{\tan }^6\frac{\mathrm{\pi }}{9}-6{\tan }^4\frac{\mathrm{\pi }}{9} \\ 3+27{\tan }^4\frac{\mathrm{\pi }}{9}-18{\tan }^2\frac{\mathrm{\pi }}{9}=9{\tan }^2\frac{\mathrm{\pi }}{9}+{\tan }^6\frac{\mathrm{\pi }}{9}-6{\tan }^4\frac{\mathrm{\pi }}{9} \\ \Rightarrow {\tan }^6\frac{\mathrm{\pi }}{9}-33{\tan }^4\frac{\mathrm{\pi }}{9}+27{\tan }^2\frac{\mathrm{\pi }}{9}=3 \\ ={\tan }^2\frac{\mathrm{\pi }}{3} \end{gathered}$$