Find value of $(\cos\frac{2\pi}{7})^ {\frac{1}{3}} + (\cos\frac{4\pi}{7})^ {\frac{1}{3}} + (\cos\frac{8\pi}{7})^ {\frac{1}{3}} $ This question was on my list. I was trying to apply the $n$-th roots of unity, but other ideas are welcome. I also tried Newton's sums, but it's not working.
I searched around here and I didn't find a similar one, but if they do, just say that I delete the topic.

This cubic-root sum, which has the inscrutable value $\sqrt[3]{(5-3\sqrt[3]{7})/2}$, was discovered over a hundred years ago. Outlined below is an elementary evaluation of the sum.
Note that $\cos\frac{2\pi}7$, $\cos\frac{4\pi}7 $ and $\cos\frac{8\pi}7 $ are the roots of
$$x^3+\frac{1}{2}x^2-\frac{1}{2}x-\frac{1}{8}=0$$
With the short-hands,
$$c_1=\left(\cos\frac{2\pi}{7}\right)^{\frac{1}{3}},\space \space \space c_2=\left(\cos\frac{4\pi}{7}\right)^{\frac{1}{3}},\space \space \space c_3=\left(\cos\frac{8\pi}{7}\right)^{\frac{1}{3}}
$$
we have
$$c_1^3+c_2^3+c_3^3 = -\frac{1}{2},\>\>\>
c_1^3c_2^3+c_2^3c_3^3+c_3^3c_1^3 = -\frac{1}{2},\>\>\>
c_1^3c_2^3c_3^3 = \frac{1}{8}\tag{1}$$
Let
$$A=c_1+c_2+c_3, \space\space\space B=c_1c_2+c_2c_3+c_3c_1$$
and evaluate
$$A^3 = c_1^3+c_2^3+c_3^3 +3AB-3c_1c_2c_3,$$
$$B^3= c_1^3c_2^3+c_2^3c_3^3+c_3^3c_1^3+3c_1c_2c_3 AB-3(c_1c_2c_3)^2
$$
Plugging (1) into above expressions to get
$$A^3=-2+3AB,\space\space\space B^3=-\frac{5}{4}+\frac{3}{2}AB\tag{2}$$
Next, evaluate
$$(2AB-3)^3=8A^3B^3-36A^2B^2+54AB-27$$
and use the results (2) to get the equation satisfied by $A$,
$$\left( \frac{2A^3-5}{3}\right)^3 =-7\tag{3}$$
Finally, solve (3) and we obtain the sum,
$$A = \left(\cos\frac{2\pi}{7}\right)^{\frac{1}{3}} +\left(\cos\frac{4\pi}{7}\right)^{\frac{1}{3}}+\left(\cos\frac{8\pi}{7}\right)^{\frac{1}{3}} = 
\left[ \frac{1}{2}\left(5-3\cdot 7^{\frac{1}{3}}\right) \right]^{\frac{1}{3}}$$
