I think we can further simplify the above boolean function (^ is for XOR):

(((A B) ^ (A + B)) (A ^ B)) =           
apply A ^ B = (A'B) + (A B')
( ((A B)' (A + B)) + ((A B) (A + B)')) (A ^ B)) =   
apply DeMorgan law (X+Y+...)'=X'Y'Z'... and (XYZ...)'=X'+Y'+...
( ((A' + B') (A + B)) + ((A B) (A' B'))) (A ^ B)) = 
apply Distributive Law X(Y+Z) = XY + XZ
(A'A + A'B + AB' + BB' + AA' + BB') (A ^ B) =       
apply X+X=X, XX=X
(A'B + AB') (A ^ B) =
(A ^ B) (A ^ B) = A ^ B = A xor B

So we get the same result by just using a simple A XOR B. Is the above simplifacation correct? Can anyone confirm that?