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.

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. 
