Show that the set $A \cap B = \emptyset$ Let  $A$ and $B$ be two sets for which the following applies: $A \cup B = (A \cap B^{C}) \cup (A^{C} \cap B)$. Show that $A \cap B = \emptyset$.
How?! I am seriously stuck. One thought I had is to distribute the right part, so that: $(A \cap B^{C}) \cup (A^{C} \cap B) = (A \cap B^{C} \cup A^{C}) \cup (A \cap B^{C} \cup B^{C})$ 

I'll add an answer expanding your idea, since that works too!
Only, you made a few mistakes.
It should be:  
$(A \cap B^{C}) \cup (A^{C} \cap B) = ((A \cap B^{C}) \cup A^{C}) \cap ((A \cap B^{C}) \cup B)$   
Then, 
$=((A\cup A^{C}) \cap (B^{C}\cup A^{C})) \cap ((A\cup B) \cap (B^{C}\cup B))=$
$=(B^{C}\cup A^{C}) \cap (A\cup B)$  
We know this to be $A\cup B$, so we can conclude $A \cap B = \emptyset$ (for instance by drawing a Venn diagram, or by further manipulation)
