Is $x+y\alpha + z\alpha^2$ a subfield? Let $\alpha = \sqrt[3]{2}$. I want to Prove that the set of all numbers $x+y\alpha+z\alpha^2$, for $x,y,z \in \mathbb{Q}$, is a subfield of $\mathbb{C}$. 
I have showed that this set is a subring, but I'm not sure how to show for any $a\neq 0$ in our set has a multiplicative inverse. 
I have tried picking two arbitrary elements, $m,n$ in our set. Let $m= x+y\alpha+z\alpha^2$ and let $n=a+b\alpha+c\alpha^2$. Then setting our product $m\cdot n$ equal to 1 gets us:
$$(ax+2zb+2yc) + (bx+ay+2zc)\alpha + (cx + yb +za)\alpha^2 = 1.$$
I was wondering if this could lead me to my proof? 

Recall the algebraic identity
$$x^3 + y^3 + z^3 -3xyz = (x+y+z)(x^2 + y^2+z^2 - xy -yz - zx)\\
\implies
\frac{1}{x+y+z} =  \frac{x^2+y^2+z^2 - xy-yz-zx}{x^3+y^3+z^3 - 3xyz}
$$
Substitute $(x,y,z)$ by $(x,y\alpha,z\alpha^2)$, one get
$$\frac{1}{x + y\alpha + z\alpha^2}
= \frac{(x^2-2yz) + (2z^2-xy)\alpha + (y^2-zx)\alpha^2}{x^3+2y^3 + 4z^3 -6xyz}$$
Aside from justifying why $x^3+2y^3+4z^3-6xyz \ne 0$, this gives you 
a formula for the multiplicative inverse of $x + y\alpha + z\alpha^2$.
