Prove or disprove $ \ (A \times A) - (B \times B) = (A-B) \times (A-B)$ Question: 
Let $ A,B$ be sets.
Prove or disprove: $ \ (A \times A) - (B \times B) = (A-B) \times (A-B)$
My attempt:
Let $ \ (x,y) \in (A \times A) - (B \times B) \implies (x,y) \in (A \times A)$ and $ \ (x,y) \notin (B \times B) \implies x \in A $ and $ \ x\notin B$ and $ \ y \in A$ and $ \ y \notin B \implies (x,y) \in (A-B) \times (A-B)$
Let $ \ (x,y) \in (A-B) \times (A-B) \implies x \in A$ and $ x \notin B$ and $ \ y \in A$ and $ \ y \notin B \implies (x,y) \in (A \times A)$ and $ \ (x,y) \notin (B \times B) \implies (x,y) \in (A \times A) - (B \times B)$.
Is this approach correct?

This is incorrect.
The set $A \times A$, is the sets where both coordinates belong in $A$.
The set $B \times B$ is the set where both coordinates belong in $B$.
So, $A \times A-B \times B$ are the points where both coordinates belong in $A$ except the ones that both coordinates belong in $B$. In other words, the first coordinates or the second coordinates does not belong in $B$.
Without words,
$$ A \times A-B \times B= \left ((A-B)\times A \right ) \cup \left (A\times (A-B)  \right )$$
