How to prove $P(A\cap B) - P(A)P(B) = P(A)P(B^c) - P(A\cap B^c)$? I'm trying to prove the identity in the title. I have tried all the results I'm aware of but I can't find the way out. I know that
$$P(A) = P(A\cap B)+ P(A\cap B^c),$$
and that
$$P(A\cup B) = P(A)+P(B)-P(A\cap B),$$
but I don't see how to combine these (or other properties I am missing) to get the desired result...
Any hint is highly appreciated.

You are almost there. Note that
$$
P(A)\cdot 1= P(A)(P(B)+P(B^c)) = P(A\cap B) + P(A\cap B^c),
$$
and the second property is not needed here. Do you see now the way out?
