1. Identidad de $\tan(3\theta)$:
$$ \tan(3\theta) = \frac{3\tan\theta - \tan^3\theta}{1 - 3\tan^2\theta} $$
Sea $\theta = \pi/9$, entonces $3\theta = \pi/3$. Sabemos que $\tan(\pi/3) = \sqrt{3}$.
$$ \sqrt{3} = \frac{3\tan(\pi/9) - \tan^3(\pi/9)}{1 - 3\tan^2(\pi/9)} $$

2. Elevando al cuadrado:
Sea $t = \tan(\pi/9)$.
$$ 3 = \left( \frac{3t - t^3}{1 - 3t^2} \right)^2 \implies 3(1 - 3t^2)^2 = (3t - t^3)^2 $$
$$ 3(1 - 6t^2 + 9t^4) = 9t^2 - 6t^4 + t^6 $$
$$ 3 - 18t^2 + 27t^4 = 9t^2 - 6t^4 + t^6 $$

3. Reordenando términos:
$$ t^6 - 27t^4 - 6t^4 + 18t^2 + 9t^2 = 3 $$
$$ t^6 - 33t^4 + 27t^2 = 3 $$
Sustituyendo $t = \tan(\pi/9)$:
$$ \boxed{\tan^6 \left( \frac{\pi}{9} \right) - 33 \tan^4 \left( \frac{\pi}{9} \right) + 27 \tan^2 \left( \frac{\pi}{9} \right) = 3} $$