I have found the minimal polynomial if $\zeta$ over $\mathbb{Q}$ is $x^{6}+x^{3}+1$. $\mathbb{Q}(\zeta)\colon\mathbb{Q}$ is a normal and separable extension so $\lvert$Gal($\mathbb{Q}(\zeta)\colon\mathbb{Q})\rvert=6$.
Let $G=$Gal$(\mathbb{Q}(\zeta)\colon\mathbb{Q})$. $G$ contains the automorphisms that send $\zeta$ to $\zeta^{i}$ where $i=1,2,4,5,7,8.$ The group is generated by the automorphism $\pi\colon\zeta\rightarrow\zeta^{2}$. So $G=\{e,\pi,\pi^{2},\pi^{3},\pi^{4},\pi^{5}\}\cong(\mathbb{Z}\backslash9\mathbb{Z})^{\times}$.
The subgroups of $G$ are $\{e\}$, $\{e,\pi^{3}\}=\langle\pi^{3}\rangle$, $\{e,\pi^2,\pi^{4}\}=\langle\pi^{2}\rangle$ and $\{e,\pi,\pi^{2},\pi^{3},\pi^{4},\pi^{5}\}=\langle\pi\rangle$. Using the Galois correspondence I found the subfields of $\mathbb{Q}(\zeta)$ which are $\mathbb{Q}(\zeta^{3})$ which has degree 2 over $\mathbb{Q}$ and $\mathbb{Q}(\zeta+\zeta^{8})$ which has degree 3 over $\mathbb{Q}$.
Now the minimal polynomial $f(x)$ of an element in $\mathbb{Q}(\zeta)$ is solvable by radicals if there is a radical extension $L$ of $\mathbb{Q}(\zeta)$ such that $L$ contains the splitting field of $f(x)$. And $f(x)$ is solvable by radicals if and only if its Galois group is solvable.
I'm not sure how to continue or if any of the previous calculations are useful in answering this question? Any help would be appreciated, thank you.
The extension $\mathbb{Q}(\zeta)$ is normal and so contains the splitting field of each $\alpha \in \mathbb{Q}(\zeta)$. Thus, the Galois group of the splitting field of any such $\alpha$ is a quotient of $(\mathbb{Z}/9 \mathbb{Z})^{\times}$ and so is Abelian. All Abelian groups are solvable.