I have seen this same question answered here, but I would like to know if I could arrive to the solution with any of my approaches. Splitting field of x^3-2 as a simple extension
Clearly $\mathbb{Q}(\sqrt[3]{2} + e^{2 \pi i/3}) \subseteq \mathbb{Q}(\sqrt[3]{2}, e^{2 \pi i/3})$, so the only thing left to do is prove it the other way.
1st attemp
To see that $\sqrt[3]{2}, e^{2 \pi i/3} \in \mathbb{Q}(\sqrt[3]{2} + e^{2 \pi i/3})$ would be enough. For this, I have seen that a common approach is to compute $(\sqrt[3]{2} + e^{2 \pi i/3})^{-1}$ and generaly you obtain an element that summed with $\sqrt[3]{2} + e^{2 \pi i/3}$ results in $\sqrt[3]{2}$ or $e^{2 \pi i/3}$. I am having a really hard time trying to rationalize it.
2nd attemp
I try to reason it by using the degrees of the extension. We have that $[\mathbb{Q}(\sqrt[3]{2}, e^{2 \pi i/3}) : \mathbb{Q}(\sqrt[3]{2} + e^{2 \pi i/3})] = \frac{[\mathbb{Q}(\sqrt[3]{2}, e^{2 \pi i/3}) : \mathbb{Q}]}{[\mathbb{Q}(\sqrt[3]{2} + e^{2 \pi i/3}) : \mathbb{Q}]}$.
I think the esiest way to get $[\mathbb{Q}(\sqrt[3]{2}, e^{2 \pi i/3}) : \mathbb{Q}]$ is by presenting a $\mathbb{Q}$-base. Although I think I am not doing it right, because seeing this question Splitting field of x^3-2 as a simple extension, I think it should be of degree 6, not 7.
$\mathbb{Q}(\sqrt[3]{2}, e^{2 \pi i/3}) = \mathbb{Q}\langle1, \sqrt[3]{2}, \sqrt[3]{4}, \sqrt[3]{2}e^{2 \pi i/3}, \sqrt[3]{2}e^{4 \pi i/3}, \sqrt[3]{4}e^{2 \pi i/3}, \sqrt[3]{4}e^{4 \pi i/3}\rangle$
To compute $[\mathbb{Q}(\sqrt[3]{2} + e^{2 \pi i/3}) : \mathbb{Q}]$ we can try to find the irreducible polynomial.
$$ \alpha = \sqrt[3]{2} + e^{2 \pi i/3} \\ \alpha^3 = (\sqrt[3]{2} + e^{2 \pi i/3})^3 \\ \alpha^3 = 2 + 3\sqrt[3]{4}e^{2 \pi i/3} + 3\sqrt[3]{2}e^{4 \pi i/3} + 1 \\ \frac{1}{3}(\alpha^3 - 3) = \sqrt[3]{4}e^{2 \pi i/3} + \sqrt[3]{2}e^{4 \pi i/3} \\ \frac{1}{3}(\alpha^3 - 3) = \sqrt[3]{2}e^{2 \pi i/3}(\sqrt[3]{2} + e^{2 \pi i/3}) \\ \frac{1}{3}(\alpha^3 - 3) = \sqrt[3]{2}e^{2 \pi i/3}\alpha \\ \frac{1}{27}(\alpha^3 - 3)^3 = 2\alpha^3 \\ \frac{1}{27}(\alpha^9 - 9\alpha^6 + 27\alpha^3-27) = 2\alpha^3 \\ \alpha^9 - 9\alpha^6 - 27\alpha^3-27 = 0 $$
So clearly, $\sqrt[3]{2} + e^{2 \pi i/3}$ is a root of the polynomial $p(x) = x^9 - 9x^6 - 27x^3-27$. But it does not seem irreducible.
I am stuck at this point. Thanks in advance, any help would be appreciated.
We have $$ x^9-9x^6-27x^3-27=(x^6 + 3x^5 + 6x^4 + 3x^3 + 9x + 9)(x^3 - 3x^2 + 3x - 3). $$ The factors are irreducible, so we have the desired result for degree $6$, which you have linked: Splitting field of x^3-2 as a simple extension