Two roots of polynomial If a polynomial with rational coefficients has a root $1 + \cos(2\pi/9) + \cos^2(2\pi/9)$, then the one also has a root $1+\cos(8\pi/9)+\cos^2(8\pi/9).$ How to prove it?

Take $\omega=e^{i2\pi/9}$.
Let $\alpha=1 + \cos(2\pi/9) + \cos^2(2\pi/9)$.
Then $4\alpha = 6+2\omega+ \omega^2 + \omega^7 + 2\omega^8$, by using $2\cos(2\pi/9)=\omega+\bar\omega$ and $\omega^9=1$.
Let $\beta = 1+\cos(8\pi/9)+\cos^2(8\pi/9)$. Then $4\beta = 6 + \omega+ + 2\omega^4 + 2\omega^5 + \omega^8$, by using $2\cos(8\pi/9)=\omega^4+\bar\omega^4$.
Now note that the map $\omega \mapsto \omega^4$ on $\mathbb Q(\omega)$ sends $\alpha$ to $\beta$ and the map $\omega \mapsto \omega^5$ sends $\beta$ to $\alpha$. This means that for $p$ a polynomial with rational coefficients, $p(\alpha)=0$ iff $p(\beta)=0$, as required.
Alternatively, you can also conclude that
$\alpha$ and $\beta$ are conjugate algebraic numbers and your result follows because their minimal polynomial is the same.
