I want to construct a cyclic division algebra of degree $3$ over some degree $3$ Galois extension $E$ of $\mathbb{Q}$. So the construction is as follows:


As a set $D=E\oplus uE \oplus u^2 E$ where $u$ is an indeterminate. Addition is defined component wise while multiplication is defined as $e.u=u\sigma(e)$ where $\sigma\in Gal(E/\mathbb{Q})$ is fixed and $u^3=a$ for some fixed $a\in \mathbb{Q}\setminus N\_{E/\mathbb{Q}}(E)$ where $N\_{E/\mathbb{Q}}$ is the norm map.


I want to find particular $a$ for some extension for some calculation purpose in $D^\*$.


I started with $E=\frac{\mathbb{Q}[x]}{x^3+x^2-2x-1}$ and calculated that 
$N\_{E/\mathbb{Q}}(a+bx+cx^2)= a^3-a^2b-2ab^2+b^3+5ac^2-abc-b^2c+6ac^2-2bc^2+c^3$. 
But it is difficult to find one element which is not norm of any element of the extension in this manner.


So here is my question:


Can you find explicitly one $a\in\mathbb{Q}$ such that $N\_{E/\mathbb{Q}}(\alpha)\neq a$ for all $\alpha\in E$.



Image of the norm map for degree $3$ galois extension over $\mathbb{Q}$
There is a general strategy to tackle such question. Since the norm is multiplicative, it make sense to look for a prime $p$ not in the image. Then, we can ask about the number of times $p$ divides a general norm. 
The norm map extends to prime ideals of the integer ring, and if the norm $N\_{E/F}(q)$ is $p^3$ for every such ideal over $p$, then $p$ can not possibly be a norm, since its ideal is even not a norm of a fractional ideal of $\mathcal{O}\_E$. This happens exactly for those primes $p$ which are inert in $E$, namely those primes modulo which the defining polynomial of $E$ remains irreducible. Indeed, the norm $N\_{E/F}(q)$ coincide the the size of $\mathcal{O}\_E/q$ which is $p^3$ for inert $q$, the quotient being $\mathbb{F}\_{p^3}$. 


In our case, for example, the polynomial $x^3+x^2-2x-1$ reduces mod $2$ to the polynomial $x^3+x^2+1$ which is irreducible, so $2$ is an inert prime and the number $2$ can not be realized as a norm. So you can choose $a=2$. 


edit: Apparently a similar answer was put in the comments while I written this, sorry for the double answer. 


