<s>[INST] Splitting field of $\alpha=(2+\sqrt{2})^{1/3}$ Given $\alpha=(2+\sqrt{2})^{1/3}$ The problem asks to calculate the minimal polynomial of $\alpha$ over $\mathbb{Q}$ and find it's splitting field. I have the solution and I understood until the point where it shows that the minimal polynomial is $$x^6-4x^3+2=0$$
Now, since we got this polynomial developing the equation $\alpha=(2+\sqrt{2})^{1/3}$ it is clear that $(2+\sqrt{2})^{1/3}$ is a root of the minimal polynomial. 
I don't understand the following: Why are $\left( 2-\sqrt{2} \right) ^{1/3},\left( 2-\sqrt{2} \right) ^{1/3}w,\left( 2-\sqrt{2} \right) ^{1/3}w^2,\left( 2+\sqrt{2} \right) ^{1/3}w,\left( 2+\sqrt{2} \right) ^{1/3}w^2
$ the other roots of the polynomial? $w$ is the cubic root of unity ($w^3=1$).
 [/INST] $$x^6-4x^3+2=0$$ $$\Leftrightarrow (x^3-2)^2=2$$
$$\Leftrightarrow x^3-2=\pm \sqrt{2}$$
$$\Leftrightarrow x^3=2\pm \sqrt{2}$$
$$\Leftrightarrow  x=\omega^i\sqrt[3]{2+\sqrt{2}},\omega^i\sqrt[3]{2-\sqrt{2}},i=0,1,2,$$ where $\omega$ is a root of $x^2+x+1$, so $\omega^3=1$.
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