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Prove that \[\begin{aligned} \tan^6 \frac{\pi}{9}+\tan^6 \frac{2\pi}{9} +\tan^6 \frac{4\pi}{9} \end{aligned}\] is an integer, and find the value of this integer.

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If \(x = \tan \dfrac{k\pi}{9}\) for any \(k \in \overline{-4,4}\). Then \(\arg(1+ix) = \dfrac{k\pi}{9}\), and so, \(\arg[(1+ix)^9] = k\pi\).

Thus, \(\text{Im}[(1+ix)^9] = 0\). So, \(x = \tan \dfrac{k\pi}{9}\) is a root of \(\text{Im}[(1+ix)^9] = 0\) for \(k \in \overline{-4,4}\).

Using the binomial theorem, \(P(x) := \text{Im}[(1+ix)^9] = x^9 - 36x^7 + 126x^5 - 84x^3 + 9x\).

Since \(\tan 0 = 0\) and \(\tan \dfrac{\pm3\pi}{9} = \pm \sqrt{3}\) are roots of \(P(x)\), we know \(x(x^2-3)\) divides \(P(x)\).

Factoring yields \(P(x) = x(x^2-3)(x^6-33x^4+27x^2-3)\).

Therefore, \(\pm \tan \dfrac{\pi}{9}\), \(\pm \tan \dfrac{2\pi}{9}\), \(\pm \tan \dfrac{4\pi}{9}\) are the roots of \(Q(x) := x^6-33x^4+27x^2-3\).

Thus, \(\tan^2 \dfrac{\pi}{9}\), \(\tan^2 \dfrac{2\pi}{9}\), \(\tan^2 \dfrac{4\pi}{9}\) are the roots of \(R(x) := Q(\sqrt{x}) = x^3-33x^2+27x-3\).

Let \(S_n\) be the sum of the \(n\)-th powers of the roots of \(R(x)\). Now, we use Newton's Sums.

\(S_1 -33 \cdot 1 = 0 \leadsto S_1 = 33\).

\(S_2 -33 \cdot S_1 + 27 \cdot 2 = 0 \leadsto S_2 = 1035\).

\(S_3 -33 \cdot S_2 + 27 \cdot S_1 -3 \cdot 3 = 0 \leadsto S_3 = 33273\).

Therefore, \(S_3 = \tan^6 \dfrac{\pi}{9} + \tan^6 \dfrac{2\pi}{9} + \tan^6 \dfrac{4\pi}{9} = \boxed{33273}\), which is clearly an integer.

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Lovely.

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## Comments

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TopNewestIf \(x = \tan \dfrac{k\pi}{9}\) for any \(k \in \overline{-4,4}\). Then \(\arg(1+ix) = \dfrac{k\pi}{9}\), and so, \(\arg[(1+ix)^9] = k\pi\).

Thus, \(\text{Im}[(1+ix)^9] = 0\). So, \(x = \tan \dfrac{k\pi}{9}\) is a root of \(\text{Im}[(1+ix)^9] = 0\) for \(k \in \overline{-4,4}\).

Using the binomial theorem, \(P(x) := \text{Im}[(1+ix)^9] = x^9 - 36x^7 + 126x^5 - 84x^3 + 9x\).

Since \(\tan 0 = 0\) and \(\tan \dfrac{\pm3\pi}{9} = \pm \sqrt{3}\) are roots of \(P(x)\), we know \(x(x^2-3)\) divides \(P(x)\).

Factoring yields \(P(x) = x(x^2-3)(x^6-33x^4+27x^2-3)\).

Therefore, \(\pm \tan \dfrac{\pi}{9}\), \(\pm \tan \dfrac{2\pi}{9}\), \(\pm \tan \dfrac{4\pi}{9}\) are the roots of \(Q(x) := x^6-33x^4+27x^2-3\).

Thus, \(\tan^2 \dfrac{\pi}{9}\), \(\tan^2 \dfrac{2\pi}{9}\), \(\tan^2 \dfrac{4\pi}{9}\) are the roots of \(R(x) := Q(\sqrt{x}) = x^3-33x^2+27x-3\).

Let \(S_n\) be the sum of the \(n\)-th powers of the roots of \(R(x)\). Now, we use Newton's Sums.

\(S_1 -33 \cdot 1 = 0 \leadsto S_1 = 33\).

\(S_2 -33 \cdot S_1 + 27 \cdot 2 = 0 \leadsto S_2 = 1035\).

\(S_3 -33 \cdot S_2 + 27 \cdot S_1 -3 \cdot 3 = 0 \leadsto S_3 = 33273\).

Therefore, \(S_3 = \tan^6 \dfrac{\pi}{9} + \tan^6 \dfrac{2\pi}{9} + \tan^6 \dfrac{4\pi}{9} = \boxed{33273}\), which is clearly an integer.

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Lovely.

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Yes

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