Show that if $\gcd(a,b)=1$, then $\gcd(a+b,a^2+b^2-ab)=1\text{ or } 3$. I come across things like the following but I don't quite know how to use them.
$$\gcd(a+b,a^2+b^2-ab)|3ab$$
$$\gcd(a,b)=1\Rightarrow \gcd(a+b,ab)=1$$

$a^2 + b^2 - ab = (a + b)^2 - 3ab$
so
$$\gcd(a+b,a^2 + b^2 - ab) = \gcd(a + b， 3ab)$$
If $3$ does not divide $ a + b$ then 
$$\gcd(a+b,a^2 + b^2 - ab) = \gcd(a + b， ab) = \gcd(a,b) = 1$$
otherwise let $a + b = 3k,\,\, k\in \mathbb{Z}$, then $b = 3k - a$ and 
$$\gcd(a+b,a^2 + b^2 - ab) = 3\gcd(k， a(3k - a)) = 3\gcd(k,a^2) = 3$$
since $3$ does not divide $a$ and $\gcd(a,k) = \gcd(a,3k) = \gcd(a,3k - a) = \gcd(a,b) = 1$.
