(A+B) (A+B') (A'+B), (A'+B')

(A+B) (A+B') (A'+B), (A'+B')

= (AA+AB+AB+BB') (A'A'+A'B'+A'B+BB')

= $(A+AB'+AB+0) (A'+A'B'+A'B+0) \ \ \ \ \ \because$ Since, A.A=1 & A'A' =A', BB'=0

= $(A+A(B+B')) (A'+A'(B'+B)) \ \ \ \ \ \ but (B'+B)=1$

= (A+A(1)) (A'+A'(1))

= (A+A) (A'+A')

$\therefore AA'+AA'+AA+AA' \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ but \ AA'= 0$

$\therefore 0+0+0+0= 0$