Showing $\sqrt[3]{\cos\frac{2\pi}{9}}+\sqrt[3]{\cos\frac{4\pi}{9}}+\sqrt[3]{\cos\frac{8\pi}{9}} = \sqrt[3]{\frac{3\sqrt[3]9-6}{2}} $ 
Show that
$$\sqrt[3]{\cos\frac{2\pi}{9}}+\sqrt[3]{\cos\frac{4\pi}{9}}+\sqrt[3]{\cos\frac{8\pi}{9}} = \sqrt[3]{\frac{3\sqrt[3]9-6}{2}} $$

I tried to find a polynomial that had its roots, but the degree grows too fast and I'm getting lost.
 A: Sketch of the proof using @saulspatz ninth degree equation above, though all computations that follow from that should be easily done by trigonometry too:
Let $a,b,c, \omega a, \omega b, \omega c, \omega^2 a, \omega^2 b, \omega^2 c$ the roots of the 9th degree equation $8x^9-6x^3+1=0$, where $a,b,c$ are the real ones, so $\sqrt[3]{\cos(\frac{2\pi}{9})},\sqrt[3]{\cos(\frac{4\pi}{9})},\sqrt[3]{\cos(\frac{8\pi}{9})}$ and $w^3=1$ is a primitive cubic root of unity. 
Let $d=a+b+c, q=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$. Clearly $a^3+b^3+c^3=0$ as it is a third of the cubic sum of the roots of the degree $9$ equation above and that has no coefficients in degrees $6,7,8$ (so by Newton formulas) and similarly $\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}=6$ using the reciprocal equation and Newton's formulas - obviously direct trigonometric proofs should be easy and available too
Now $p=abc=-\frac{1}{2}$ (the product of the 9 roots is $-\frac{1}{8}$ and $abc$ is the real cube root of that by inspection) or again by trigonometry.
Using $d^3=a^3+b^3+c^3+3abcdq-3abc$ and the same equation for $q^3=\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}+3\frac{dq}{abc}-3\frac{1}{abc}$, we get the following system in $d,q$:
$2d^3+3dq-3=0$, 
$q^3+6dq-12=0$
Now this leads to a cubic in $d^3$ or we can "see" that the only real roots are $d^3=\frac{3\sqrt[3]9-6}{2}, q^3=6(1+\sqrt[3]9), dq=3-\sqrt[3]9$ using the OP's identity
A: We make use of the general equality that $$\sf a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)\tag1$$ where $\sf a,b,c=\sqrt[3]{\cos\frac{2\pi}9},\sqrt[3]{\cos\frac{4\pi}9},\sqrt[3]{\cos\frac{8\pi}9}$ such that $\sf a^3+b^3+c^3=0$ and $\sf abc=-\frac12$. 
We can also substitute in that $\sf a^2+b^2+c^2=(a+b+c)^2-2(ab+bc+ca)$ where simplifying gives: $$\sf \frac32=(a+b+c)\left((a+b+c)^2-3(ab+bc+ca)\right)$$ and so $$\sf \frac32=p(p^2-3q)\tag2$$ where $\sf p=a+b+c$ and $\sf q=ab+bc+ca$.
Next we can also substitute $\sf (a,b,c)\mapsto(ab,bc,ca)$ in $\sf(1)$ to get that $$\sf a^3b^3+b^3c^3+c^3a^3-3a^2b^2c^2=(ab+bc+ca)(a^2b^2+b^2c^2+c^2a^2-a^2bc-ab^2c-abc^2)$$ such that $\sf a^3b^3+b^3c^3+c^3a^3=-\frac34$ and $\sf a^2b^2c^2=\frac14$. 
Now substitute $\sf a^2b^2+b^2c^2+c^2a^2=(ab+bc+ca)^2-2abc(a+b+c)$ to get: $$\sf -\frac32=(ab+bc+ca)\left((ab+bc+ca)^2-3abc(a+b+c)\right)$$ and so since $\sf abc=-\frac12$, $$\sf -\frac32=q(q^2+\frac32p)\tag3$$
Thus we get a system of cubics $\sf(2)$ & $\sf(3)$. 
We can solve by first canceling out the $\sf pq$ terms to get $$\sf q^3=-\frac12p^3-\frac34$$ so that $\sf(2)$ becomes $$\sf \left(p^3-\frac32\right)^3=27p^3\left(-\frac12p^3-\frac34\right)$$
This is a cubic in $\sf p^3$, so we can proceed through the usual way and get that $$\sf p=\sqrt[3]{\frac{3\sqrt[3]9-6}2}$$ as desired. 
