Prove that $\tan^6 20°+\tan^6 40°+\tan^6 80°$ is an integer Prove that $\tan^6 20°+\tan^6 40°+\tan^6 80°$ is an integer. Doesn't this problem seem a little out of the box? It seems beautiful, but I don't have an idea on how to start. Calculating the value does give me an integer-$32733$. Thanks.

Let $\theta = 20^\circ = \frac{\pi}{9}$ and let $\omega_k = \tan(k\theta)$ for $k = 0,1,\ldots 8$. The sum we want can be rewritten as 
$$\tan^6(20^\circ) + \tan^6(40^\circ) + \tan^6(80^\circ) = \omega_1^6 + \omega_2^6 + \omega_4^6$$
Since
$(\cos(k\theta) + i\sin(k \theta))^9 = e^{i9k\theta} = (-1)^k$, 
the $\omega_k$ are the nine roots of the polynomial
$$
\Im\left[( 1 + i t )^9\right] 
= t^9-36t^7+126t^5-84t^3+9t
= t(t^2-3)(t^6-33t^4+27t^2-3)
$$
Notice $\omega_0 = 0, \omega_3 = -\omega_6 = \sqrt{3}$, the factor
$t(t^2 - 3) = (t - \omega_0)(t - \omega_3)(t - \omega_6)$.
Together with the identities $\omega_1 = -\omega_8$, $\omega_2 = -\omega_7$, $\omega_4 = -\omega_5$, we find
$$(x - \omega_1^2)(x - \omega_2^2)(x - \omega_4^2) = x^3 - 33x^2 + 27 x - 3$$
Let $p_n = \omega_1^{2n} + \omega_2^{2n} + \omega_3^{2n}$ for $n \in \mathbb{Z}_{+}$. Apply Newton's identities to above cubic polynomial, we have
$$
\begin{cases}
p_1 - 33 = 0\\
p_2 - 33 p_1 + 2\times 27 = 0\\
p_3 - 33 p_2 + 27 p_1 - 3\times 3 = 0
\end{cases}
\implies 
\begin{cases}
p_1 = 33\\
p_2 = 33 \times 33 - 2\times 27 = 1035\\
p_3 = 33 \times 1035 - 27 \times 33 + 3\times 3 = 33273
\end{cases}
$$
As a result,
$$\tan^6(20^\circ) + \tan^6(40^\circ) + \tan^6(80^\circ) = \omega_1^6 + \omega_2^6 + \omega_4^6 = p_3 = 33273$$
