Every cubic extension $L/K$ is of the shape $L=K(\sqrt[3]{\alpha})$ with $\alpha$ not a cubed number and $\text{char}(K)\neq3$ I want to disprove this but I don't really know how to proceed from my 'answer' below where $|\beta|=3$; why is $|\beta|=3$ true when $\mathbb{F}_2$={0,1} (if it is even true in the first place). In other words, why can we say: $\beta^3 = 1$ implies that $|\beta| \mid 3$?
 A: We will give a counterexample to show that not every cubic extension $L/K$ is of the shape $L=K(\sqrt[3]{\alpha})$ with $\alpha$ not a cubed number and $\text{char}(K)\neq3$. Let $K = \mathbb{F}_2$, $L = K[X]/(X^3+X+1)$ and $f:=X^3+X+1$. This function $f$ is irreducible for $f(0)=1\neq 0$ and $f(1)=3=1\neq0$. Hence $L$ is a field, and $L/K$ is a field extension of degree 3; i.e. a cubic extension. Now suppose that $L=K(\sqrt[3]{\alpha})$, for some $\alpha \in K$. 
Then it must be true that we have a $\beta \in L\backslash K$ with $\beta^3 \in K$. 
But if $\beta^3 \in K$ then either $\beta^3 = 0$ or $\beta^3 = 1$. 
But $\beta^3 = 0$ implies $\beta = 0$ since a field has no zero divisors. 
And $\beta^3 = 1$ implies that $|\beta| \mid 3$, but then either $|\beta| = 1$ or $|\beta| = 3$. In the case where $\beta = 1 \in K$ we have a contradiction. 
In the case where $|\beta|=3$, we also have a contradiction since the order of the element has to divide the order of the (multiplicative) group which is $|\mathbb{F}_2^{\times}|=8-1=7$. 
Hence, such a $\beta$ does not exist.$\Box$.  
Edit
A way easier counterexample is the field $K=\mathbb{Q}\subset\mathbb{Q}(\cos(2\pi/7))=L$. The element $\cos(2\pi/7)\notin\mathbb{Q}$ has the following minimum polynomial: $f=8X^3+4X^2-4X-1\in\mathbb{Q}[X]$. We know that $\deg_x(f)=3$, hence, $[L:K]=3$.
Proof for this minimum polynomial of $\cos(2\pi/7)$. We know that $\cos(2\pi/7)=\Re(e^{\frac{2i\pi}{7}})=:\alpha$ which is the real part of the 7-th root of unity $\zeta_7$ with minimum polynomial $\Phi_7(X)=X^6+X^5+X^4+X^3+X^2+X+1$.
Let $p:=2\pi/7$. We now compute the following: $$\Phi_7(\alpha)=\cos(6p)+\cos(5p)+\cos(4p)+\cos(3p)+\cos(2p)+\cos(p)+1.$$
We compute each term separately:
$\cos(6p)=\cos(12\pi/7)=\cos(2\pi-2\pi/7)=\cos(2\pi/7)=\cos(p).$
$\cos(5p)=\cos(10\pi/7)=\cos(2\pi-2\cdot2\pi/7)=\cos(2p).$
$\cos(4p)=\cos(8\pi/7)=\cos(\pi+\pi/7)=-\cos(\pi/7).$
$\cos(3p)=\cos(6\pi/7)=\cos(\pi-\pi/7)=-\cos(\pi/7)\rightarrow \cos(3p)=\cos(4p)$.
$\cos(3p)=4\cos^3(p)-3\cos(p)$ (formula of De Moivre for $n=3$).
$\cos(2p)=2\cos^2(p)-1$.
Now we rewrite $\Phi_7(\alpha)$ as follows:
$$\Phi_7(\alpha)=\cos(p)+2\cos^2(p)-1+2(4\cos^3(p)-3\cos(p))+2\cos^2(p)-1+\cos(p)+1=$$
$$8\cos^3(p)+4\cos^2(p)-4\cos(p)-1=8\alpha^3+4\alpha^2-4\alpha-1.$$
