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? 

You have to solve
$$
\begin{cases}
ax+2bz+2yc=1\\
ay+bx+2cz=0\\
az+by+cx=0
\end{cases}
$$
in the variables $a,b,c$. The notation here is somewhat of counterintuitive with respect to the usual one.
The matrix associate to the non-homogeneous linear system is
$$
A(x,y,z)=\begin{pmatrix}
x & 2z & 2y\\
y & x & 2z\\
z & y & x
\end{pmatrix}
$$
and the determinant is $x^3+2y^3+4z^3-6xyz$. Assuming $A(x,y,z)$ invertible, you can determine $(a,b,c)$ as $\displaystyle A(x,y,z)^{-1}\cdot \begin{pmatrix} 1\\0\\0 \end{pmatrix}$.
It remains to show that $\det A(x,y,z)=x^3+2y^3+4z^3-6xyz=0$ iff $(x,y,z)=(0,0,0)$.
Set $X=x$, $Y=y\sqrt[3]{2}$ and $Z=z\sqrt[3]{4}$. 
Hence the equality above becomes: $x^3+2y^3+4z^3-6xyz=X^3+Y^3+Z^3-3XYZ=0$
It is known that 
$$X^3+Y^3+Z^3-3XYZ=\frac12(X+Y+Z)\big((X-Y)^2+(Y-Z)^2+(Z-X)^2\big)$$
hence we get: 
$X^3+Y^3+Z^3-3XYZ=\frac12(X+Y+Z)\big((X-Y)^2+(Y-Z)^2+(Z-X)^2\big)=0$.
There are two possible cases:


*

*$X+Y+Z=x+y\sqrt[3]{2}+z\sqrt[3]{4}=0$. Hence $x=y=z=0$ because $x,y,z\in \Bbb Q$. You can easily check that $1,\sqrt[3]{2},\sqrt[3]{4}$ are linearly independent over $\Bbb Q$.

*$\big((X-Y)^2+(Y-Z)^2+(Z-X)^2\big)=0$. Since this is a sum of three squares, the total amount is zero iff each summand in zero, that is $(X-Y)=(Y-Z)=(Z-X)=0$ iff $X=Y=Z$ iff $x=y\sqrt[3]{2}=z\sqrt[3]{4}=0$ iff $x=y=z=0$ because $\sqrt[3]{2},\sqrt[3]{4}\notin\Bbb Q$.
Therefore: $\det A(x,y,z)=0$ iff $x=y=z=0$, iff $m=x+y\alpha+z\alpha^2=0$. Hence any element different to zero has a multiplicative inverse.
