Norm in the cyclotomic integers 
Let $\alpha \in \mathbb{Z}[\zeta_3]$, where $\zeta_3=e^{2\pi i/3}$ is a cube root of unity. So $\alpha=x+y\zeta_3$ for $x,y\in\mathbb{Z}$. Show that the norm $N(\alpha)$ can be written as $\frac{a^2+3b^2}{4}$ for integers $a,b$ of the same parity.

I think this is supposed to be a somewhat straightforward arithmetic-heavy problem, but I'm having trouble with it.
Since $\zeta_3^2=-1-\zeta_3$, I was able to show that $N(\alpha)=x^3+y^3$, but I don't see how this reduces to what I need.

The norm $x^2+y^2-xy$ can be simply computed via square-completion procedure to get
$4(x^2+y^2-xy)$=$(x^2+2xy+y^2)+3(x^2-2xy+y^2)$
.              =$(x+y)^2+3(x-y)^2$
And also since
$x^2+y^2-xy=x^2+(x^2-2xy+y^2)-(x^2-xy)$
.            =$x^2+(x-y)^2-x(x-y)$
The we also have
$4(x^2+y^2-xy)=(2x-y)^2+3y^2$
Either way, it shows that the norm can be written in the form $a^2+3b^2$, by dividing the above expressions $4$ depending on whether$ x$ and $y$ are of the same parity or not. Use the first expression when $x$ and $y$ are of the same parity and use the second expression when $x$ and $y$ are of different parity, assuming $y$ is even and $x$ is odd.
