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 ($w^3=1$).
 A: $$x^6-4x^3+2=0$$ $$\Leftrightarrow (x^3-2)^2=2$$
$$\Leftrightarrow x^3-2=\pm \sqrt{2}$$
$$\Leftrightarrow x^3=2\pm \sqrt{2}$$
$$\Leftrightarrow  x=\omega^i\sqrt[3]{2+\sqrt{2}},\omega^i\sqrt[3]{2-\sqrt{2}},i=0,1,2,$$ where $\omega$ is a root of $x^2+x+1$, so $\omega^3=1$.
A: First, remember that if $\alpha$ is a (real or complex) root of $x^n-a=0$, $a\neq 0$, then the other roots are all of the form $\lambda^i\alpha$, where $\lambda$ is a primitive $n$th root of unity. That is, a complex number such that $\lambda^n=1$ but $\lambda^k\neq 1$ for $1\leq k\leq n-1$. This follows simply by noting that all of them are roots of $x^n-a$, they are all distinct, since $\lambda^k\alpha = \lambda^j\alpha$ implies $\lambda^k=\lambda^j$, and therefore that $\lambda^u=1$ where $u=|k-j|$. That means $n|k-j$, and so you get equality if both $k$ and $j$ are between $0$ and $n-1$.
Now, your polynomial is 
$$x^6 - 4x^3 + 2.$$
This is a quadratic polynomial in $x^3$. So you can easily find the two values of $x^3$ that are roots. Call them $a_1$ and $a_2$. So then you want to solve $x^3=a_i$, or equivalently find a root of $x^3-a_i$, $i=1,2$. 
Since $Y^2 - 4Y + 2= 0$ has roots $2+\sqrt{2}$ and $2-\sqrt{2}$, then you want to find the solutions to
$$x^3 - (2+\sqrt{2})\qquad\text{and}\qquad x^3-(2-\sqrt{2}).$$
One root of the first is $\sqrt[3]{2+\sqrt{2}}$; so the three roots are $\sqrt[3]{2+\sqrt{2}}$, $\omega\sqrt[3]{2+\sqrt{2}}$, and $\omega^2\sqrt[3]{2+\sqrt{2}}$. 
Similarly, since $\sqrt[3]{2-\sqrt{2}}$ is a root of the second polynomial, the other two roots are $\omega\sqrt[3]{2-\sqrt{2}}$ and $\omega^2\sqrt[3]{2-\sqrt{2}}$. 
