Problem source

Show, using de Moivre's theorem, or otherwise, that 
\[
\tan9\theta=\frac{t(t^{2}-3)(t^{6}-33t^{4}+27t^{2}-3)}{(3t^{2}-1)(3t^{6}-27t^{4}+33t^{2}-1)},\qquad\mbox{ where }t=\tan\theta.
\]
By considering the equation $\tan9\theta=0,$ or otherwise, obtain a cubic equation with integer coefficients whose roots are 
\[
\tan^{2}\left(\frac{\pi}{9}\right),\qquad\tan^{2}\left(\frac{2\pi}{9}\right)\qquad\mbox{ and }\qquad\tan^{2}\left(\frac{4\pi}{9}\right).
\]
Deduce the value of 
\[
\tan\left(\frac{\pi}{9}\right)\tan\left(\frac{2\pi}{9}\right)\tan\left(\frac{4\pi}{9}\right).
\]
Show that 
\[
\tan^{6}\left(\frac{\pi}{9}\right)+\tan^{6}\left(\frac{2\pi}{9}\right)+\tan^{6}\left(\frac{4\pi}{9}\right)=33273.
\]

Solution source

Writing $c = \cos \theta, s = \sin \theta$ then de Moivre states that:

\begin{align*}
&& \cos 9 \theta + i \sin 9 \theta &= (c +i s)^9 \\
&&&= c^9 + 9ic^8s - 36c^7s^2-84ic^6s^3+126c^5s^4 + 126ic^4s^5 -84c^3s^6 -36ic^2s^7+9cs^8+is^9 \\
&&&= (c^9-36c^7s^2+126c^5s^3-84c^3s^6+8cs^8)+i(9c^8s-75c^6s^3+126c^4s^5-36c^2s^7+s^9) \\
\Rightarrow && \tan 9\theta &= \frac{(9c^8s-75c^6s^3+126c^4s^5-36s^2c^7+s^9)}{(c^9-36c^7s^2+126c^5s^4-84c^3s^6+8cs^8)} \\
&&&= \frac{9t-75t^3+126s^5-36t^7+t^9}{1-36t^2+126t^4-84t^6+8t^8} \\
&&&= \frac{t(t^{2}-3)(t^{6}-33t^{4}+27t^{2}-3)}{(3t^{2}-1)(3t^{6}-27t^{4}+33t^{2}-1)}
\end{align*}

If we consider $\tan 9\theta = 0$ it will have the roots $\theta = \frac{n \pi}{9}, n \in \mathbb{Z}$, in particular, the numerator of our fraction for $\tan 9 \theta$ will be zero for $t = 0, \tan \frac{\pi}{9}, \tan \frac{2\pi}{9}, \tan \frac{3\pi}{9}, \tan \frac{4 \pi}{9}, \tan \frac{5\pi}{9}, \tan \frac{6 \pi}{9}, \tan \frac{7 \pi}{9}, \tan \frac{8\pi}{9}$. It's worth noting all other values of $\theta$ will repeat these values. Also note that $0,\tan \frac{\pi}{3}, \tan \frac{2\pi}{3}$ are the roots of $t$ and $t^2-3$ respectively. Therefore the other values are the roots of our sextic. However, also note that $\tan \frac{8\pi}{9} = - \tan \frac{\pi}{9}$ and similar, therefore we can notice that all the roots in pairs can be mapped to $\tan \frac{\pi}{9}, \tan \frac{2 \pi}{9}$ and $\tan \frac{4 \pi}{9}$ and all those values are squared, so the roots of:

$x^3 - 33x^2+27x-3$ will be $\tan^2 \frac{\pi}{9}, \tan^2 \frac{2 \pi}{9}$ and $\tan^2 \frac{4 \pi}{9}$.

The product of the roots will be $3$, so 

\begin{align*}
&& \tan^2 \frac{\pi}{9} \tan^2 \frac{2 \pi}{9} \tan^2 \frac{4 \pi}{9} &= 3 \\
\Rightarrow && \tan \frac{\pi}{9} \tan \frac{2 \pi}{9} \tan \frac{4 \pi}{9} &= \pm \sqrt{3} \\
\underbrace{\Rightarrow}_{\text{all positive}} && \tan \frac{\pi}{9} \tan \frac{2 \pi}{9} \tan \frac{4 \pi}{9} &=  \sqrt{3} \\
\end{align*}

Notice that $x^3 + y^3 +z^3 - 3xyz = (x+y+z)((x+y+z)^2-3(xy+yz+zx))$

Therefore
\begin{align*}
\tan^{6}\left(\frac{\pi}{9}\right)+\tan^{6}\left(\frac{2\pi}{9}\right)+\tan^{6}\left(\frac{4\pi}{9}\right) &= 33(33^2-3\cdot27) + 3 \cdot 3 \\
&= 33\,273
\end{align*}