# Finding the splitting field of $f(x)$

I'm trying to learn the theory of splitting fields. So I went through this example on an old exam: Find the splitting field $K$ of $f(x)$ over $\mathbb{Q}$ for $f(x)=x^6-9$

$x^6-9=(x^3-3)(x^3+3)$ and so $\pm\sqrt[3]{3}$ is the only real roots. When defining $w=-\frac{1}{2}+\frac{\sqrt[3]{3}}{2} i$ we have that $w$ is a root of $x^2+x +1$ and therefor a root of $x^{3}$$-$$1$ and so $\sqrt[3]{3}$$w$ is a solution and so is $\sqrt[3]{3}w^{2}$ since $w^3=1$ and $(w^2)^3=1$ and the same argument goes for the negative real root and so $K=\mathbb{Q}(\sqrt[3]{3},w)$ Is the argument above correct? Or am I missing something? Is this how you always go about when finding the splitting field? Finding a real root and then multiplying with roots of unity? What if there are no real roots? I'm doing this algebra course on my own and field extensions and splitting fields is a part of it that I don't get. So please explain as basic as possible and if you know any online lectures on the subject please recommend them to me.

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Yes, your argument is correct (except that in the formula for $w$, you want $\sqrt3$, not $\root3\of3$). No, that method won't always work. It works for "binomials", that is, polynomials of the form $x^n-a$. Different polynomials require different methods, some are much harder than others, some really can't be done at all until you get to the main theorems of Galois Theory (and even then, the answers may not be in the form you'd like). Keep on studying, you'll develop more tools as you go along.