GCD of cubic polynomials I would appreciate some help finding $GCD(a^3-3ab^2, b^3-3ba^2)$; $a,b \in \mathbb{Z}$. So far I've got here: if $GCD(a,b)=d$ then $\exists  \alpha, \beta$ so that $GCD(\alpha, \beta)=1$ and $\alpha d=a$, $\beta d=b$.
Therefore we know that $GCD(a^3-3ab^2, b^3-3ba^2)=d^3 GCD(\alpha^3-3\alpha \beta^2, \beta^3-3\beta \alpha^2)$. However I don't know to figure out $GCD(\alpha^3-3\alpha \beta^2, \beta^3-3\beta \alpha^2)$ given that $GCD(\alpha,\beta)=1$.

I'll write $\alpha=m,\beta=n$ for the ease of typing
If $d(\ge1)$ divides $m^3-3mn^2,n^3-3m^2n$
$d$ will divide $n(m^3-3mn^2)+3m(n^3-3m^2n)=-8m^3n$
and $3n(m^3-3mn^2)+m(n^3-3m^2n)=-8mn^3$
Consequently, $d$ must divide $(-8m^3n,-8mn^3)=8mn(m^2,n^2)=8mn$
So, $d$ will divide $8$
As $(m,n)=1,$ both $m,n$ cannot be even
If $m$ is even, $n^3-3m^2n$ will be odd $\implies d=1$
If both $m,n$ are odd,. $m^2,n^2\equiv1\pmod8$
$m(m^2-3n^2)\equiv m(1-3)\equiv-2m\pmod8$
Similarly, we can establish the highest power of $2$ in $m^3-3mn^2$ will be $2$
$\implies d=2$ if $m,n$ are odd
