Building a cubic function with integer coefficients and trigonometric roots

I want to find the answer to the following problem: Construct a cubic polynomial with integer coefficients, whose roots - $\cos{\frac{2 \pi}{7}}$, $\cos{\frac{4 \pi}{7}}$ and $\cos{\frac{6 \pi}{7}}$.

I have the following idea. These trigonometric roots are the extremum points of the Chebyshev polynomial $T_7 (x) = T_7 (\cos{t}) = \cos{7t}$ and $T_7(x_k) = 1$, where $x_k$ required the roots of a cubic function (see problem). Then the 7 degree polynomial $T_n(x)-1$ will have roots $x_k$. But the problem is that we need a polynomial of degree 3. I'm sure these trigonometric roots will be associated with the Chebyshev polynomials.What do you think about this ? Can you suggest your ideas to solve?

• I'm confused... you want a cubic equation with specific roots. There is only $one$ such equation... so it either has integer coefficients or it doesn't, there's no constructing... – Asier Calbet Sep 23 '15 at 22:58
• @Assaultous2 Only one, up to multiplication by non-zero scalars :) – Inactive - avoiding CoC Sep 23 '15 at 23:09
• You are right. This will be a polynomial $(x-\cos{\frac{2 \pi}{7}})(x-\cos{\frac{4 \pi}{7}})(x-\cos{\frac{6 \pi}{7}})$, and the problem reduces to the difficult trigonometric calculations. I think this problem can be solved easier, using Chebyshev polynomials. – Victor Sep 23 '15 at 23:15
• @ Servaes of course – Asier Calbet Sep 24 '15 at 10:11

How about the polynomial $$8\left(x-\cos\tfrac{2\pi}{7}\right)\left(x-\cos\tfrac{4\pi}{7}\right)\left(x-\cos\tfrac{6\pi}{7}\right)=8x^3+4x^2-4x-1?$$
• Sorry, I made a mistake in the description of the problem. There should be 3 different roots $\cos{\frac{2 \pi}{7}}$, $\cos{\frac{4 \pi}{7}}$ and $\cos{\frac{6 \pi}{7}}$. – Victor Sep 23 '15 at 23:09