The problem says: Given the irreducible polynomial $x^3-3x-1$ with root $2\cos(\pi/9)$, prove that $2\cos(\pi/27)$ is a root of a monic irreducible polynomial of degree 9 over $\mathbb{Q}$, and hence $[\mathbb{Q}(2\cos(\pi/27)):\mathbb{Q}]=9.$
To solve it, I substituted the equality $2\cos(\pi/9)=8\cos^3(\pi/27)-6\cos(\pi/27)$ on the equation $x^3-3x-1=0$ to obtain that $2\cos(\pi/27)$ is a root of the monic polynomial $x^9-9x^7+27x^5-30x^3+9x-1$. But now, how I prove that $x^9-9x^7+27x^5-30x^3+9x-1$ is irreducible over $\mathbb{Q}$?
The computation is a little horrendous (I used WolframAlpha), but you can show that $$(x+1)^9-9(x+1)^7+27(x+1)^5-30(x+1)^3+9(x+1)-1$$ expands to be $$x^9+9 x^8+27 x^7+21 x^6-36 x^5-54 x^4+9 x^3+27 x^2-3$$ which is irreducible by Eisenstein's criterion with $p=3$, which means that $$x^9-9x^7+27x^5-30x^3+9x-1$$ is also irreducible.
EDIT:
As Michalis points out, we can avoid the computation just by considering the polynomial modulo $3$, and then considering the constant coefficient modulo $9$. Since we have coefficients which are multiples of $3$, $P(x+1)\pmod{3}$ becomes $$(x+1)^9-1\pmod3\equiv x^9+1-1\pmod3\equiv x^9\pmod3$$ Hence all the coefficients of the expanded polynomial are divisible by $3$ except for the first. Now we consider the constant coefficient modulo $9$: This is just $$1-9+27-30+9-1\pmod9=1-0+0-3+0-1\pmod9\equiv-3\pmod9$$ So $9$ doesn't divide the final coefficient, and we have satisfied all the conditions necessary to use Eisenstein's criterion as above.