how to show that $\mathbb{Q}[\sqrt[3]{2}]$ is a field? (by elementary means) To be very concrete, I want to show that every element of the form 
$1/(p+qx+rx^2)$ where $x=\sqrt[3]{2}$ where $p,q,r$ are rationals can be written in the form $a+bx+cx^2$ where again $a,b,c$ are integers. 
Clearly, it is enough to show that certain 3x3 matrix with coefficients involving $p,q,r$ has nonzero determinant. However, I cannot show why it cannot be zero for rational values of $p,q,r$. Any simple way to do this?
 A: I assume the determinant you want is
$$\det\pmatrix{p&2r&2q\cr q&p&2r\cr r&q&p\cr}=p^3+2q^3+4r^3-6pqr\ .$$
The determinant obviously is zero when $p=q=r=0$; suppose that this is not the case and the determinant is still zero, $p,q,r$ being rational.  Multiplying by a common denominator and then cancelling any common factor, we may assume that $p,q,r$ are integers with no common factor.  If
$$p^3+2q^3+4r^3-6pqr=0$$
then $2\mid p^3$ so $2\mid p$, say $p=2s$.  Substituting back and cancelling $2$ gives
$$4s^3+q^3+2r^3-6qrs=0\ .$$
This shows that $2\mid q$; then by a similar process you get $2\mid r$; this contradicts the fact that $p,q,r$ have no common factor.  Therefore the determinant cannot be zero for rational $p,q,r$, not all zero.
I am sure you can see the similarity between this proof and the standard irrationality proof for $\sqrt2\,$.
It is actually easier to do this by using the fact that $x^3-2$ is irreducible, but you asked for a determinant proof so that's what you've got ;-)
