What comes after $\cos\left(\tfrac{2\pi}{7}\right)^{1/3}+\cos\left(\tfrac{4\pi}{7}\right)^{1/3}+\cos\left(\tfrac{6\pi}{7}\right)^{1/3}$?

Solution 1:

As some people said, the 2nd formula is easy to derive. In Maple, there are commands to get the minimal polynomial of LHS of the 2nd. The following method works for the 2nd formula, but not for the 3rd.

Prove that $\sqrt[3]{\cos\frac{2\pi}{7}} + \sqrt[3]{\cos\frac{4\pi}{7}} + \sqrt[3]{\cos\frac{6\pi}{7}} = \sqrt[3]{\frac{5-3\sqrt[3]{7}}{2}}$.

$\textit{Proof}$: $\cos\frac{2\pi}{7},\cos\frac{4\pi}{7},\cos\frac{6\pi}{7}$ are the three distinct real roots of $8x^3+4x^2-4x-1=0$. Let $u_1=\sqrt[3]{\cos\frac{2\pi}{7}}, u_2=\sqrt[3]{\cos\frac{4\pi}{7}}, u_3=\sqrt[3]{\cos\frac{6\pi}{7}}$ which are all real. All roots of $8u^9+4u^6-4u^3-1=0$ are $u_1, t u_1, t^2 u_1, u_2, t u_2, t^2 u_2, u_3, t u_3, t^2 u_3$ where $t = e^{2\pi i/3}$.

Let $(u-u_1)(u-u_2)(u-u_3)=u^3+a_2u^2+a_1u - 1/2$. We have \begin{align} (u-tu_1)(u-tu_2)(u-tu_3) &=u^3+ta_2u^2+t^2a_1u - 1/2, \\ (u-t^2u_1)(u-t^2u_2)(u-t^2u_3) &=u^3+t^2a_2u^2+ta_1u - 1/2. \end{align} It follows from \begin{align} &8u^9+4u^6-4u^3-1 \\ =\ & 8(u^3+a_2u^2+a_1u - 1/2)(u^3+ta_2u^2+t^2a_1u - 1/2)(u^3+t^2a_2u^2+ta_1u - 1/2) \end{align} that \begin{align} a_2^3-3a_1a_2-2 &=0, \\ 4a_1^3+6a_1a_2+5 &= 0. \end{align} From these two equations, we get $a_1 = \frac{a_2^3-2}{3a_2}$ and $4a_2^9+30a_2^6+75a_2^3-32=0$. Let $f(v)=4v^9+30v^6+75v^3-32$. We have $f'(v)=9v^2(2v^3+5)^2\ge 0$. Thus, $f(v)=0$ has a unique real root. It is easy to check that $v=-\sqrt[3]{\frac{5-3\sqrt[3]{7}}{2}}$ is a root of $f(v)=0$. Thus, $a_2 = -\sqrt[3]{\frac{5-3\sqrt[3]{7}}{2}}$. Thus, $u_1+u_2+u_3 = -a_2 = \sqrt[3]{\frac{5-3\sqrt[3]{7}}{2}}$. This completes the proof.

In general, suppose $x_1,x_2,x_3$ are the three distinct real roots of $x^3+bx^2+cx+d^3=0$, where $b,c,d$ are rational numbers. We have \begin{equation} \sqrt[3]{x_1}+\sqrt[3]{x_2}+\sqrt[3]{x_3} = \sqrt[3] {-\frac{3}{2}\Big(\sqrt[3]{Q+4\sqrt{\Delta}}+\sqrt[3]{Q-4\sqrt{\Delta}}\Big) -b-6d }, \end{equation} where $Q = 24bd^2+36d^3+4bc+24cd, \Delta = -4b^3d^3-27d^6+18bcd^3+b^2c^2-4c^3$. Here, $\Delta$ is the discriminant of $x^3+bx^2+cx+d^3=0$.

For the 3rd, similarly, $x_k = \cos\frac{2k\pi}{11}, k=1,\cdots, 5$ are the five distinct real roots of $32x^5+16x^4-32x^3-12x^2+6x+1=0$. Let $u_1 = \sqrt[5]{\cos\frac{2\pi}{11}}, u_2= \sqrt[5]{\cos\frac{4\pi}{11}}, u_3 = \sqrt[5]{\cos\frac{6\pi}{11}}, u_4 = \sqrt[5]{\cos\frac{8\pi}{11}}, u_5= \sqrt[5]{\cos\frac{10\pi}{11}}$. All roots of $32u^{25}+16u^{20}-32u^{15}-12u^{10}+6u^5+1=0$ are $t^j u_k, k=1,2,3,4, 5; j=0,1,2,3,4$ where $t= e^{2\pi i/5}$.

Let $\prod_{k=1}^5 (u-u_k) = u^5 + a_4u^4+a_3u^3+a_2u^2+a_1u+\frac{1}{2}$. Similarly, we have four equations of \begin{align} 0 & = a_4^5-5a_3a_4^3+5a_2a_4^2+5a_3^2a_4-5a_1a_4-5a_2a_3+2, \\ 0 & = 8a_1^5-20a_1^3a_2+10a_1^2a_3+10a_1a_2^2-5a_1a_4-5a_2a_3+1,\\ 0 & = -10a_1a_2a_4^3+10a_1a_3^2a_4^2+10a_2^2a_3a_4^2-10a_2a_3^3a_4+2a_3^5+10a_1^2a_4^2-10a_1a_2a_3a_4\\ &\quad-10a_1a_3^3-10a_2^3a_4+10a_2^2a_3^2-5a_3a_4^3+10a_1^2a_3+10a_1a_2^2+10a_2a_4^2+10a_3^2a_4\\ &\quad -15a_1a_4-15a_2a_3+7,\\ 0 & = -40a_1^3a_3a_4+40a_1^2a_2^2a_4+40a_1^2a_2a_3^2-40a_1a_2^3a_3+8a_2^5-40a_1^3a_2+20a_1^2a_4^2\\ &\quad-20a_1a_2a_3a_4-20a_1a_3^3-20a_2^3a_4+20a_2^2a_3^2+40a_1^2a_3+40a_1a_2^2+10a_2a_4^2\\ &\quad +10a_3^2a_4-30a_1a_4-30a_2a_3+13. \end{align}

However, it seems no hope for determination of $a_4$.