<s>[INST] How can I prove that $\mathbb{Q}(\sqrt[3]{2},i)=\mathbb{Q}(\sqrt[3]{2}+i)$? It is easy to show that $\mathbb{Q}(\sqrt[3]{2}+i)\subseteq \mathbb{Q}(\sqrt[3]{2},i)$. 
But how I can show that $\mathbb{Q}(\sqrt[3]{2},i)\subseteq\mathbb{Q}(\sqrt[3]{2}+i)$?
I can't find a way to express $\sqrt[3]{2}$ in terms of $\sqrt[3]{2}+i$.
 [/INST] Follows a sequence of maneuvers which shows how to express $i$ and $\sqrt[3]2$ in terms of $\sqrt[3]2 + i$, and hence that $\Bbb Q(\sqrt[3]2, i) = \Bbb Q(\sqrt[3]2 + i)$:
$\alpha = \sqrt[3]2 + i; \tag 1$
$\alpha - i = \sqrt[3]2; \tag 2$
$(\alpha - i)^3 = 2; \tag 3$
$\alpha^3 - 3 \alpha^2 i - 3 \alpha + i = 2; \tag 4$
$i(1 - 3\alpha^2) = 2 + 3\alpha - \alpha^3; \tag 5$
$i = \dfrac{\alpha^3 - 3\alpha - 2}{3\alpha^2 - 1} = \dfrac{2 +  3\alpha - \alpha^3}{1 - 3\alpha^2}  \in \Bbb Q(\alpha) = \Bbb Q(\sqrt[3]2 + i); \tag 6$
$\sqrt[3]2 = \alpha - i \in \Bbb Q(\alpha) = \Bbb Q(\sqrt[3]2 + i); \tag 7$
from (6), (7) and the fact that $\Bbb Q(\alpha) = \Bbb Q(\sqrt[3]2 + i) \subset \Bbb Q(\sqrt[3]2, i)$ we affirm
$\Bbb Q(\sqrt[3]2, i) \subset \Bbb Q(\alpha) \Longrightarrow \Bbb Q(\sqrt[3]2, i) = \Bbb Q(\alpha) = \Bbb Q(\sqrt[3]2 + i). \tag 8$
N.B. We may in fact derive from (6) a sixth degree polynomial satisfied by $\alpha$:
$\alpha^3 - 3\alpha - 2 = i(3\alpha^2 - 1), \tag 9$
$(\alpha^3 - 3\alpha - 2)^2 = -(3\alpha^2 - 1)^2, \tag{10}$
$\alpha^6 + 9\alpha^2 + 4 - 6\alpha^4 - 4\alpha^3 + 6\alpha = -9\alpha^4 + 6\alpha^2 - 1, \tag{11}$
$\alpha^6 + 3\alpha^4 - 4\alpha^3 + 3\alpha^2 + 6\alpha + 1 = 0, \tag{12}$
consistent with the results of Parcly Taxel.
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