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?

$\gcd(a+b, a^2 - ab + b^2) = \gcd(a+b, (a+b)^2 - 3ab) = \gcd(a+b, 3ab)$
Since $\gcd(a,b) = 1$ hence $\gcd(a+b,a) = 1$ and $\gcd(a+b, b)$ = 1.
Thus, $\gcd(a+b, 3ab) = \gcd (a+b, 3) $ by applying the Lemma below.
Hence, $\gcd(a+b, a^2 -ab + b^2) = \gcd(a+b, 3)=1$ or 3. (This is very simple to evaluate for given values.)

Lemma If $\gcd(m,n) = 1$, then $\gcd (m, np) = \gcd(m,p)$.
Proof: This is obvious when you consider GCD via prime factorization.
