If $\gcd(a,b)=1$, then $\gcd(a+b,a^2 -ab+b^2)=1$ or $3$. Hint:  $a^2 -ab +b^2 = (a+b)^2 -3ab.$
I know we can say that there exists an $x,y$ such that $ax + by = 1$.  So in this case,
$(a+b)x + ((a+b)^2 -3ab)y =1.$
I thought setting $x = (a+b)$ and $y = -1$ would help but that gives me $3ab =1.$  Any suggestions?

Hint $\,\ c\mid a\!+\!b,\overbrace{a^2\!-\!ab\!+\!b^2}^{\large f(b)}\Rightarrow\,c\mid 3a^2,3ab,3b^2\Rightarrow\,c\mid(3a^2,3ab,3b^2)=3(a,b)^2 = 3 $ 
by $\,\ {\rm mod}\ \ a\!+\!b\!:\ \ \color{#c00}{b\equiv -a}\,\Rightarrow\, f(\color{#c00}b)\equiv f(\color{#c00}{-a})\equiv 3a^2\equiv 3a(-b)\equiv 3(-b)(-b)  $ 
