<s>[INST] $\cos \left(\frac{2\pi }{7}\right)^{\frac{1}{3}}+\cos \left(\frac{4\pi }{7}\right)^{\frac{1}{3}}+\cos \left(\frac{6\pi }{7}\right)^{\frac{1}{3}} =?$ I have been trying to solve the following question for a long time:
Find $a,b,c,d$ such that: ($a,b,c,d$ are primes)

$\cos \left(\frac{2\pi }{7}\right)^{\frac{1}{3}}+\cos \left(\frac{4\pi }{7}\right)^{\frac{1}{3}}+\cos \left(\frac{6\pi }{7}\right)^{\frac{1}{3}} = \left(\frac{a-b\sqrt[3]{c}}{d}\right)^{\frac{1}{3}} $

What I have tried so far is that:
Let $x=\cos \left(\frac{2\pi }{7}\right), y=\cos \left(\frac{4\pi }{7}\right),z=\cos \left(\frac{6\pi }{7}\right)$, then
$x+y+z=-1/2$,
$xyz =1/8 $
$xy+yz+zx=-1/2$
So, $x,y,z$ are the roots of $8t^3+4t^2-4t-1=0$ and can be found by Cardan's method and hence $x^{1/3}+y^{1/3}+z^{1/3}$ can be found. However that method was way too long and it seems that the question may have a more elegant solution (given the form).
Also, I tried to cube both sides of the equation in the question and substitute using the above three relations, but that required the value of $x^{2/3}+y^{2/3}+z^{2/3}$, which I coundn't find.
Would someone pls help me out (without using Cardan)?
Thanks in advance!
 [/INST] Try Ramanujan's cubic polynomial, this polynomial don't need to find 3 roots. If coefficients are satisfied the condition, then sum of cuberoots of zeroes can be evaluated directly from its coefficients.
EDIT: With your results, you established the equation: $8t^3+4t^2-4t-1=0$ which is equivalent $t^3+\frac{1}{2}t^2-\frac{1}{2}t-\frac{1}{8}=0$
The Ramanujan's cubic polynomial said that: If the polynomial $x^3+px^2+qx+r$ has 3 real roots $x_1, x_2, x_3$ and the relation $pr^{\frac{1}{3}}+3r^{\frac{2}{3}}+q=0$ then $x_1^{\frac{1}{3}}+ x_2^{\frac{1}{3}}+ x_3^{\frac{1}{3}}=\left(-p-6r^{\frac{1}{3}}+3(9r-pq)^{\frac{1}{3}}\right)^{\frac{1}{3}}$
Back to our case: $p=\frac{1}{2}, q=-\frac{1}{2}, r=-\frac{1}{8}$ and $pr^{\frac{1}{3}}+3r^{\frac{2}{3}}+q=\left(\frac{1}{2}\right)\left(-\frac{1}{8}\right)^{\frac{1}{3}}+3\left(-\frac{1}{8}\right)^{\frac{2}{3}}-\frac{1}{2}=-\frac{1}{4}+\frac{3}{4}-\frac{1}{2}=0$ then you can calculate the sum of cuberoot of zeroes follow the relation with $x_1, x_2, x_3$. Thanks for reading.
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