As AG means “on every step of every infinite path”, we can use the well known conversion:
AGa ∧ AG(Fa ∧ ¬b) = AGa ∧ AGFa ∧ AG¬b
As a implies Fa, AGa also implies AGFa (intuitively: whenever we satisfy AGa, we also satisfy AGFa, so it makes no difference in the conjunction). Hence, we just drop AGFa. And there you go, a beautifully hand-crafted:
AGa ∧ AG¬b