boolean algebra simplification, maybe a typo in the book?

Question

I have found the following exercise:

I have tried to solve it, but in any way that I try I cannot get the answer. I did the following:

(C+A)(¬A+B)(C+B) Conmutative in terms (1) and (3)

(C+AB)(¬A+B) Distributive

So it says that A.B=0 therefore I will have:

(C+0)(¬A+B)

C(¬A+B)

BC+¬AC Distributive again

I have also tried a workaround by applying distributive to terms (2) and (3), but I end up with the same answer. Am I missing something?

Thanks

Answer 1

Hint: Multiply your result $C(\lnot A+B)$ by $1=A+B$.

By the way, the hypothesis implies $B=\lnot A$.