Finding minimal polynomial of $(\sqrt[3]{2}-2)^2$ over $\mathbb{Q}$

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I am trying to find the minimal polynomial of $\alpha= (\sqrt[3]{2} -2)^2$. I know that $\alpha \in \mathbb{Q}(\sqrt[3]{2} ) $ and since $\alpha $ is not rational I know that the degree of the minimal polynomial must be $3$. But actually trying to manipulate it seems difficult. Are there any tips or a way to do this generally?

EDIT

I have $\alpha =\sqrt[3]{2^2}-4\sqrt[3]{2} +4$

$\alpha^2= 24\sqrt[3]{2^2}-30\sqrt[3]{2} $

and $\alpha^3= 216\sqrt[3]{2^2}-72\sqrt[3]{2}-252 $.

Now just need to combine somehow. Is there an easier strategy than this though?

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$\alpha=\left(\sqrt[3]2-2\right)^2=\left(2-\sqrt[3]2\right)^2>0$

$\sqrt\alpha=2-\sqrt[3]2>0$

$\sqrt\alpha-2=-\sqrt[3]2$

$\left(\sqrt\alpha-2\right)^3=-2$

$\alpha\sqrt\alpha-6\alpha+12\sqrt\alpha-8=-2$

$(\alpha+12)\sqrt\alpha=6(\alpha+1)$

$\alpha(\alpha+12)^2=36(\alpha+1)^2$

$\alpha^3-12\alpha^2+72\alpha-36=0$