I was going through Bezout's identity and I had a question. Can we simplify $gcd(a*b,c+b*(d-e))$ using Bezout's identity? When I use the identity $gcd(a+m*b,b)=gcd(a,b)$ I am left with $gcd(a*b,c+b*(d-e))=gcd(a*b,c)$. Similarly, $gcd(a*b,c+b*(d+e))=gcd(a*b,c)$ i.e. both seem to be the same. But if I substitute with numbers to verify these identities, both give different answers i.e. $gcd(a*b,c+b*(d+e))\neq gcd(a*b,c+b*(d-e))$. Can someone please point out when I am going wrong?
Thanks