Greatest Common Divisor Problem: Prove that $\gcd(\frac{a^3+b^3}{a+b}, a+b) = \gcd(a+b, 3ab)$ I've been stuck in this problem for some time now. Currently what I have accomplished is, using the propriety $\gcd(a,b) = \gcd(b,a \bmod(b))$  to get in the equation
$$ \gcd(a+b, \frac{a^3+b^3}{a+b}\bmod(a+b)) $$
but I don't know where to go anymore or even if I'm in the correct path. Any tips or solutions would be great appreciated.

We have
$$
\begin{pmatrix}
\frac{a^3+b^3}{a+b} \\ a+b 
\end{pmatrix}
=
\begin{pmatrix}
a^2-ab+b^2 \\ a+b 
\end{pmatrix}
=
\begin{pmatrix}
-1 & a+b \\ 0 & 1
\end{pmatrix}
\begin{pmatrix}
3ab \\ a+b
\end{pmatrix}
$$
The result follows because the matrix has determinant $-1$ and so has an integer inverse.
