For a complex number $\alpha $ which is algebraic over $\Bbb Q$, determining whether $\bar{\alpha}\in \Bbb Q(\alpha)$ or not

Consider the extension $K = \mathbb{Q}(\sqrt[3]{3}, \sqrt[4]{3}, i)$. This is a normal extension of $\mathbb{Q}$, because it is the splitting field of $(X^{3}-3)(X^{4}-3)$. This is evident from the fact that the primitive third roots of unity are $(-1 \pm i\sqrt{3})/2$, which clearly belong to $K$. Hence, $K$ is a Galois extension of $\mathbb{Q}$. Its degree over $\mathbb{Q}$ is $24$, since $\mathbb{Q}(\sqrt[3]{3}, \sqrt[4]{3})$ has degree $12$ over $\mathbb{Q}$ (it contains subextensions of degree $3$ and $4$), and adjoining $i$ to any subfield of $\mathbb{R}$ gives an extension of degree $2$.

Moreover, we can list the elements of $\mathrm{Gal}(K/\mathbb{Q})$ explicitly in terms of the generators of $K$ over $\mathbb{Q}$. Indeed, $\sqrt[3]{3}$ must go to a root of $X^{3}-3$; $\sqrt[4]{3}$ must go to a root of $X^{4}-3$; and $i$ must go to a root of $X^{2}+1$. This gives $24$ possible automorphisms of $K$ over $\mathbb{Q}$, all of which must be realized since $[K:\mathbb{Q}] = 24$.

Back to the problem at hand, if $\overline{\alpha} \in \mathbb{Q}(\alpha)$, then $\sqrt[3]{3}$ belongs to $\mathbb{Q}(\alpha)$. From here, it is not hard to see that $i\sqrt[4]{3}$ belongs to $\mathbb{Q}(\alpha)$, and so one may deduce that $\mathbb{Q}(\alpha) = \mathbb{Q}(i\sqrt[12]{3})$. On the other hand, if we have this equality of fields, then clearly $\overline{\alpha} \in \mathbb{Q}(\alpha)$.

We always have the containment $\mathbb{Q}(\alpha) \subset \mathbb{Q}(i\sqrt[12]{3})$, and $\mathbb{Q}(i\sqrt[12]{3})$ has degree $12$ over $\mathbb{Q}$ because $i\sqrt[12]{3}$ is a root of $X^{12}-3$, so it suffices to show that $\mathbb{Q}(\alpha)$ has degree $12$ over $\mathbb{Q}$. One can do so by counting the distinct Galois conjugates of $\alpha$ by the elements of $\mathrm{Gal}(K/\mathbb{Q})$ enumerated above, and one sees that there are indeed 12 unique conjugates.

Let $\omega = 3^{1/12} i$. Note that $\omega^{12} - 3 = 0$, and $x^{12} - 3$ is irreducible over $\mathbb Q$. Now $\alpha = \omega^4 - \omega^{15}$. The resultant of $x^{12}-3$ and $x^4 - x^{15}-y$ with respect to $x$ is $$P(y) = {y}^{12}-12{y}^{9}-729{y}^{8}+54{y}^{6}-34992{y}^{5}+177147{ y}^{4}-108{y}^{3}-65610{y}^{2}-2125764y-14348826 $$ which is irreducible over $\mathbb Q$. This is the minimal polynomial of $\alpha$. It has degree $12$. Thus both $\mathbb Q(\alpha)$ and $\mathbb Q(\omega)$ are vector spaces of dimension $12$ over $\mathbb Q$. Since $\alpha \in \mathbb Q(\omega)$, $\mathbb Q(\alpha) \subseteq \mathbb Q(\omega)$, but since their dimensions are equal they must be equal. In particular, $\overline{\alpha} = \omega^4 + \omega^{15} \in \mathbb Q(\omega) = \mathbb Q(\alpha)$.