How to prove that $P(A) P(B)=P(A \cap B) P(A \cup B)+P\left(A \cap B^{\prime}\right) P\left(B \cap A^{\prime}\right)$

Question

What I tried to do so far is replace $P(B)$ with $P(B) = P(A \cap B) + P(B \cap A^c)$. With which I get, $P(A)P(A \cap B) + P(A)P(B \cap A^c)$. After that I also tried doing the same trick for $P(A)$ but it did not work out. Any help would be greatly appreciated.

Answer 1

Let $p(A\cap B’)=x$, $p(A\cap B)=y$ and $p(B\cap A’)=z$

Then $$p(A)p(B)=(x+y)(y+z)=xy+xz+y^2+yz$$ $$=y(x+y+z)+xz$$ $$=p(A\cap B)p(A\cup B)+p(A\cap B’)p(B\cap A’)$$

Answer 2

$$P(A \cap B) P(A \cup B)+P\left(A \cap B^{\prime}\right) P\left(B \cap A^{\prime}\right)=$$ $$=(P(A)+P(B)-P(A \cup B))P(A \cup B) + (P(A \cup B)-P(B))(P(A \cup B)-P(A))=$$ $$=P(A)P(A \cup B)+P(B)P(A \cup B)-(P(A \cup B))^2 + (P(A \cup B))^2 -P(A)P(A \cup B)-P(B)P(A \cup B)+P(A)P(B)=$$ $$=P(A)P(B)$$

$P\left(A \cap B^{\prime}\right)=P(A \cup B)-P(B)$

$P\left(B \cap A^{\prime}\right)=P(A \cup B)-P(A)$

$P(A \cap B)=P(A)+P(B)-P(A \cup B)$