Find an example for a field extension of $F/K$ and suppose that $[F: K] =3$. But F $\ne K(\sqrt[3]{a}) \forall a \in K$ Find an example for a field extension of $F/K$ and suppose that $[F:K]$=3. But F $\ne K(\sqrt[3]{a}) \forall a \in K$.
Thanks a lot.
 A: $\mathbb{F}_{27}$ is an extension of degree $3$ of $\mathbb{F}_3$. Since $a^3=a$ for $a \in \mathbb{F}_3$, we have $\mathbb{F}_3(\sqrt[3]{a})=\mathbb{F}_3$ for $a \in \mathbb{F}_3$.
A: An extension of the form $\mathbb Q(\sqrt[3]a)$ where $ a\in \mathbb Q$  is not the cube of a rational number cannot be normal (=Galois) because it does not contain the two other roots of the irreducible polynomial $X^3-a$, since these roots are not real.
So, contrapositively, any normal extension of $\mathbb Q$ of degree $3$ is guaranteed to be not of the form $\mathbb Q(\sqrt[3]a)$.   
But do such normal extensions exist?
Yes, they do but they are not so easy to find. Here are two examples: 
The first example is the splitting field of the polynomial $X^3+X^2-2X-1$.
If $\alpha$ is one of its roots then the other two are $\alpha^2-2$ and $\alpha^3-3\alpha$, so that indeed the extension $\mathbb Q(\sqrt[3]a)$ is normal and thus:
The degree $3$ extension $\mathbb Q(\alpha)$  is not  of the form $\mathbb Q(\sqrt[3]a)$  for any $a\in\mathbb Q $ 
The second example is obtained by considering the cyclotomic extension of level $9$ of $\mathbb Q$, namely  $K= \mathbb Q(\omega)$ where $\omega =\exp(\frac {2i\pi}{9})$.
The extension $K/\mathbb Q$ is   Galois  of degree $6$, and has cyclic Galois group.
It thus has  a unique subextension $E$ of degree $3$, automatically normal.
That extension is $\mathbb Q(\omega +\omega^{-1})$ and since $\omega +\omega^{-1}=2\cos (\frac {2\pi}{9})$ we conclude:
The degree $3$ extension $\mathbb Q( \cos (\frac {2\pi}{9}) ) $ is not of the form $\mathbb Q(\sqrt[3]a)$ for any $a\in\mathbb Q $
A: As stated the answer is NO for trivial reasons. Not only taking cubic roots of elements which are already cubes in $K$ (e.g. $0$, $1$, $8$, $27$ $...\in\Bbb Q$) gives no extension but also taking cubic roots of different non cubes yields different extensions, e.g. $3^{1/3}\notin\Bbb Q(2^{1/3})$.
A more interesting question would be: Suppose that $F\supset K$ is a cubic extension of fields of characteristic $0$ (positive characteristic is ruled out like in Martin Brandenburg answer), is it then true that $F$ can be obtained by extending $K$ with the cubic root of some $a\in K$?
The question is justified by the fact that the answer of the corresponding question for quadratic field is affirmative.
Yet, the answer is stil no. Suppose $F\supset\Bbb Q$ is a cubic Galois extension. We know that $F=\Bbb Q(\alpha)$ where $\alpha$ is the root of some irreducible polynomial $P(X)\in\Bbb Q[X]$ of degree 3. This polynomial cannot be of the form $X^3-q$ since these polynomials have 1 real root and two complex conjugate roots whereas a Galois extension of degree 3 of $\Bbb Q$ admits only real embeddings.
