<s>[INST] Find the cubic equation of $x=\sqrt[3]{2-\sqrt{3}}+\sqrt[3]{2+\sqrt{3}}$ Find the cubic equation which has a root $$x=\sqrt[3]{2-\sqrt{3}}+\sqrt[3]{2+\sqrt{3}}$$
My attempt is 
$$x^3=2-\sqrt{3}+3\left(\sqrt[3]{(2-\sqrt{3})^2}\right)\left(\sqrt[3]{(2+\sqrt{3})}\right)+3\left(\sqrt[3]{(2-\sqrt{3})}\right)\left(\sqrt[3]{(2+\sqrt{3})^2}\right)+2-\sqrt{3}$$
$$x^3=4+\left(\sqrt[3]{(2-\sqrt{3})^2}\right)\left(\sqrt[3]{(2+\sqrt{3})}\right)+3\left(\sqrt[3]{(2-\sqrt{3})}\right)\left(\sqrt[3]{(2+\sqrt{3})^2}\right)$$
then what I will do??
 [/INST] Probably it is better to notice, for first, that the product between $2-\sqrt{3}$ and $2+\sqrt{3}$ is one. After that, the minimal polynomial of $a=\sqrt[3]{2+\sqrt{3}}$ over $\mathbb{Q}$ is quite trivially:
$$ p(x)=(x^3-2)^2-3 = x^6-4x^3+1 $$
hence $a$ is a root of:
$$ x^3+x^{-3}-4 = \left(x+\frac{1}{x}\right)^3 - 3\left(x+\frac{1}{x}\right)-4$$
and $a+\frac{1}{a}$ is a root of:
$$ q(x) = x^3-3x-4.$$
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