Can you take as given the fact that cos(2 pi/7) is the real part of
a root of the 7th cyclotomic polynomial? That is,

x^6 + x^5 + x^4 + x^3 + x^2 + x + 1 = 0

If so, then consider

cos(6y) + cos(5y) + ... + cos(y) + 1 = 0

For the specific value y = 2 pi/7,

cos(6y) = cos(y)
cos(5y) = cos(2y)
cos(4y) = cos(3y)
cos(3y) = 4 (cos y)^3 - 3 cos(y)
cos(2y) = 2 (cos y)^2 - 1

Letting x = cos(2 pi/7),

x + (2 x^2 - 1) + 2 (4 x^3 - 3x) + (2 x^2 - 1) + x + 1 = 0

Collecting like terms,

8 x^3 + 4 x^2 - 4x - 1 = 0

So (2 pi/7) is indeed a root of your equation, and thus also of the
minimal polynomial x^3 + x^2 - 2x - 1.