Tan-fastic

This solution utilizes several lesser known trig identities, so even though it's very messy it may be informative.

$\color{#D61F06}{\tan x = \frac{\sin x}{\cos x}}$ $\tan^{6}\frac{\pi}{9}+\tan^{6}\frac{2\pi}{9}+\tan^{6}\frac{4\pi}{9}=\frac{\sin^{6}\frac{\pi}{9}}{\cos^{6}\frac{\pi}{9}}+\frac{\sin^{6}\frac{2\pi}{9}}{\cos^{6}\frac{2\pi}{9}}+\frac{\sin^{6}\frac{4\pi}{9}}{\cos^{6}\frac{4\pi}{9}}$ $=\frac{\color{#3D99F6}{(\sin\frac{\pi}{9}\cos\frac{2\pi}{9}\cos\frac{4\pi}{9})^{6}}\color{#333333}+\color{#EC7300}{(\sin\frac{2\pi}{9}\cos\frac{\pi}{9}\cos\frac{4\pi}{9})^{6}}\color{#333333}+\color{#20A900}{(\sin\frac{4\pi}{9}\cos\frac{\pi}{9}\cos\frac{2\pi}{9})^{6}}}{(\cos\frac{\pi}{9}\cos\frac{2\pi}{9}\cos\frac{4\pi}{9})^{6}}$ Knowing the value of the denominator first will make the chaos of finding the value of the numerator make more sense.

$\color{#D61F06}{\cos\frac{\pi}{9}\cos\frac{2\pi}{9}\cos\frac{4\pi}{9}=\frac{1}{8}}$ $\left(\cos\frac{\pi}{9}\cos\frac{2\pi}{9}\cos\frac{4\pi}{9}\right)^{6}=\left(\frac{1}{8}\right)^{6}=\frac{1}{8^{6}}$ Finding the value of the numerator (the way I did it) involves taking that mess and turning it into a larger and larger mess until it finally collapses on itself. I will show the details of the blue term. The details of the orange and green terms will follow the exact same logic and look almost identical, so I will just show the highlights for the orange and green terms.

$\color{#3D99F6}{\text{Blue Term:}}$

$\color{#D61F06}{\cos x\cos y=\frac{1}{2}[\cos(x-y)+\cos(x+y)]}$

$\color{#D61F06}{\cos x\sin y=\frac{1}{2}[\sin(x+y)-\sin(x-y)]}$

$\color{#D61F06}{\cos\frac{2\pi}{3}=-\frac{1}{2} \quad\quad \sin\frac{\pi}{3}=\frac{\sqrt{3}}{2}}$ $\left(\sin\frac{\pi}{9}\cos\frac{2\pi}{9}\cos\frac{4\pi}{9}\right)^{6}=\left(\sin\frac{\pi}{9}\left[\frac{1}{2}\left(\cos\frac{2\pi}{9}+\cos\frac{2\pi}{3}\right)\right]\right)^{6}$ $=\left(\frac{1}{2}\cos\frac{2\pi}{9}\sin\frac{\pi}{9}-\frac{1}{4}\sin\frac{\pi}{9}\right)^{6}=\left(\frac{1}{2}\left[\frac{1}{2}\left(\sin\frac{\pi}{3}-\sin\frac{\pi}{9}\right)\right]-\frac{1}{4}\sin\frac{\pi}{9}\right)^{6}$ $=\left(\frac{\sqrt{3}}{8}-\frac{1}{2}\sin\frac{\pi}{9}\right)^{6}$ $\text{Binomial Expasion}$ $=\frac{27}{8^{6}}+6\cdot \frac{9\sqrt{3}}{8^{5}}\cdot \left(-\frac{1}{2}\sin\frac{\pi}{9}\right)+15\cdot \frac{9}{8^{4}}\cdot \left(\frac{1}{4}\sin^{2}\frac{\pi}{9}\right)+20\cdot \frac{3\sqrt{3}}{8^{3}}\cdot \left(-\frac{1}{8}\sin^{3}\frac{\pi}{9}\right)$ $+15\cdot \frac{3}{8^{2}}\cdot \left(\frac{1}{16}\sin^{4}\frac{\pi}{9}\right)+6\cdot \frac{\sqrt{3}}{8}\cdot \left(-\frac{1}{32}\sin^{5}\frac{\pi}{9}\right)+\frac{1}{64}\sin^{6}\frac{\pi}{9}$ $\text{Power Reduction}$

In this part, I will combine power reduction and putting all terms in a common denominator for each of the seven terms separately and then combine then together at the end.

$\color{#D61F06}{\text{n is odd:}\quad\sin^{n}x=\frac{1}{2^{n-1}}\sum_{k=0}^{\frac{n-1}{2}}(-1)^{(\frac{n-1}{2}-k)}\left(\begin{array}{c}n\\ k\end{array}\right)\sin((n-2k)x)}\quad\color{#333333}\left(\begin{array}{c}n\\ k\end{array}\right) \text{refers to binomial expansion}$

$\color{#D61F06}{\text{n is even:}\quad\sin^{n}x=\frac{1}{2^{n}}\left(\begin{array}{c}n\\ \frac{n}{2}\end{array}\right)+\frac{1}{2^{n-1}}\sum_{k=0}^{\frac{n}{2}-1}(-1)^{(\frac{n}{2}-k)}\left(\begin{array}{c}n\\ k\end{array}\right)\cos((n-2k)x)}$

1) $\quad\frac{27}{8^{6}}\quad\text{Already good}$

2) $\quad\cdot\frac{4}{4}=\frac{-216\sqrt{3}\sin\frac{\pi}{9}}{8^{6}}$

3) $\quad=\frac{135}{8^{4}}\cdot\frac{1}{4}\cdot\frac{1-\cos\frac{2\pi}{9}}{2}\cdot\frac{8}{8}=\frac{1080-1080\cos\frac{2\pi}{9}}{8^{6}}$

4) $\quad=\frac{-60\sqrt{3}}{8^{4}}\cdot\frac{3\sin\frac{\pi}{9}-\sin\frac{\pi}{3}}{4}\cdot\frac{16}{16}=\frac{-2880\sqrt{3}\sin\frac{\pi}{9}+1440}{8^{6}}$

5) $\quad=\frac{45}{8^{3}}\cdot\frac{1}{2}\cdot\frac{3-4\cos\frac{2\pi}{9}+\cos\frac{4\pi}{9}}{8}\cdot\frac{32}{32}=\frac{4320-5760\cos\frac{2\pi}{9}+1440\cos\frac{4\pi}{9}}{8^{6}}$

6) $\quad=\frac{-6\sqrt{3}}{8^{2}}\cdot\frac{1}{4}\cdot\frac{10\sin\frac{\pi}{9}-5\sin\frac{\pi}{3}+\sin\frac{5\pi}{9}}{16}\cdot\frac{64}{64}=\frac{-3840\sqrt{3}\sin\frac{\pi}{9}+2880-384\sqrt{3}\sin\frac{5\pi}{9}}{8^{6}}$

7) $\quad=\frac{1}{8^{2}}\cdot\frac{10-15\cos\frac{2\pi}{9}+6\cos\frac{4\pi}{9}-\cos\frac{2\pi}{3}}{32}\cdot\frac{128}{128}=\frac{1344-1920\cos\frac{2\pi}{9}+768\cos\frac{4\pi}{9}}{8^{6}}$

Add and combine like terms, and finally: $\color{#3D99F6}{\text{Blue Term}}\color{#333333}=\frac{-6936\sqrt{3}\sin\frac{\pi}{9}-8760\cos\frac{2\pi}{9}+2208\cos\frac{4\pi}{9}-384\sqrt{3}\sin\frac{5\pi}{9}+11091}{8^{6}}$ $\color{#EC7300}{\text{Orange Term:}}\color{#333333}\quad\left(\sin\frac{2\pi}{9}\cos\frac{\pi}{9}\cos\frac{4\pi}{9}\right)^{6}=\left(\frac{1}{2}\sin\frac{2\pi}{9}-\frac{\sqrt{3}}{8}\right)^{6}$ $=\frac{-6936\sqrt{3}\sin\frac{2\pi}{9}-8760\cos\frac{4\pi}{9}+2208\cos\frac{8\pi}{9}-384\sqrt{3}\sin\frac{10\pi}{9}+11091}{8^{6}}$ $\color{#20A900}{\text{Green Term:}}\color{#333333}\quad\left(\sin\frac{4\pi}{9}\cos\frac{\pi}{9}\cos\frac{2\pi}{9}\right)^{6}=\left(\frac{1}{2}\sin\frac{4\pi}{9}+\frac{\sqrt{3}}{8}\right)^{6}$ $=\frac{6936\sqrt{3}\sin\frac{4\pi}{9}-8760\cos\frac{8\pi}{9}+2208\cos\frac{16\pi}{9}+384\sqrt{3}\sin\frac{20\pi}{9}+11091}{8^{6}}$ Now the blue, orange and green terms are added together. All of the trig terms cancel each other out and all that's left are the numbers. It's just hard to see because they cancel in sets of three not sets of two like you might be looking for. You might notice the blue, orange and green terms have the same set of coefficients. Since they are being added, if you group the terms with the same coefficient together, factor out that coefficient, and use these trig identities, they will indeed cancel out.

$\color{#D61F06}{\cos\frac{\pi}{9}-\cos\frac{2\pi}{9}-\cos\frac{4\pi}{9}=0}$

$\color{#D61F06}{\sin\frac{\pi}{9}+\sin\frac{2\pi}{9}-\sin\frac{4\pi}{9}=0}$

$\color{#D61F06}{\sin(\pi-x)=\sin x\quad\sin(\pi+x)=-\sin x\quad\sin(2\pi+x)=\sin x}$

$\color{#D61F06}{\cos(\pi-x)=-\cos x\quad\cos(-x)=\cos x\quad\cos(2\pi+x)=\cos x}$

1) $\quad -6936\sqrt{3}(\sin\frac{\pi}{9}+\sin\frac{2\pi}{9}-\sin\frac{4\pi}{9})= -6936\sqrt{3}(0)=0$

2) $\quad -8760(\cos\frac{2\pi}{9}+\cos\frac{4\pi}{9}+\cos\frac{8\pi}{9})=-8760(\cos\frac{2\pi}{9}+\cos\frac{4\pi}{9}\color{#3D99F6}{-\cos\frac{\pi}{9}})$

$\quad =\color{#3D99F6}{8760}\color{#333333}(\cos\frac{\pi}{9}-\cos\frac{2\pi}{9}-\cos\frac{4\pi}{9})=8760(0)=0$

3) $\quad 2208(\cos\frac{4\pi}{9}+\cos\frac{8\pi}{9}+\cos\frac{16\pi}{9})=2208(\cos\frac{4\pi}{9}\color{#3D99F6}{-\cos\frac{\pi}{9}}\color{#333333}+\color{#20A900}{\cos\frac{-2\pi}{9}})$

$\quad =2208(\cos\frac{4\pi}{9}-\cos\frac{\pi}{9}\color{#333333}+\color{#20A900}{\cos\frac{2\pi}{9}}\color{#333333})=\color{#3D99F6}-2208\color{#333333}(\cos\frac{\pi}{9}-\cos\frac{2\pi}{9}-\cos\frac{4\pi}{9})=-2208(0)=0$

4) $\quad -384\sqrt{3}(\sin\frac{5\pi}{9}+\sin\frac{10\pi}{9}-\sin\frac{20\pi}{9})=-384\sqrt{3}(\color{#3D99F6}{\sin\frac{4\pi}{9}}\color{#20A900}{-\sin\frac{\pi}{9}}\color{#333333}-\color{#69047E}{\sin\frac{2\pi}{9}})$

$\quad =\color{#3D99F6}{384\sqrt{3}}\color{#333333}(\sin\frac{\pi}{9}+\sin\frac{2\pi}{9}-\sin\frac{4\pi}{9})=384\sqrt{3}(0)=0$

$\frac{\frac{11091+11091+11091}{8^{6}}}{\frac{1}{8^{6}}}=\color{#3D99F6}{33273}\color{#333333}\quad\Box$

Considering the imaginary part of $(\cos(k\pi/9)+i\sin(k\pi/9))^9$ , we obtain the formula ${9\choose9}x^9-{9\choose7}x^7+{9\choose5}x^5-{9\choose3}x^3+{9\choose1}x=0$ for $x=\tan\left(\frac{k\pi}{9}\right)$ . Dividing out $x(x^3-3)$ , accounting for 0 and $\pm\frac{\pi}{3}$ , we end up with the even polynomial $x^6-33x^4+27x^2-3$ whose roots are $x=\tan\left(\frac{k\pi}{9}\right)$ for $k=\pm{1},\pm{2},\pm{4}$ . Thus the roots of $x^3-33x^2+27x-3$ are $x=\tan^2\left(\frac{k\pi}{9}\right)$ for $k=1,2,4$ . Now, by Girard's formula, the sum of the cubes of the roots of this polynomial is $33^3-3*33*27+3*3=\boxed{33273}$ .