# Simplify ((A ∪ B) ∩ C) ∪ ((C − A) − B).

So to find the simplest form I made an arbitrary sets A,B,C and then put them into the expression.

In the end I believe this simplifies to just C and i'm fairly confident this is the correct answer.

I was hoping you could offer a more correct or common way of simplifying this.

It should be just $$C$$. To see this, notice that $$(A\cup B) \cap C$$ means the element in C and also in A or B, and $$(C-A)-B$$ means the element in C but not in A or in B. Thus the union of the previous statement means every element in C (in A or B, or not in A or B).
You can write $$((C-A)-B) = C - (A \cup B)$$ and hence obtain that your original expression is equivalent to $$(C - (A \cup B)) \cup (C \cap (A \cup B).$$ If you think about it for a moment, you'll realize that this is of the form $$(C-D) \cup (C \cap D),$$ where $$D = A \cup B,$$ and this is just $$C$$ as you predicted.
Let $$D=((A \cup B) \cap C) \cup ((C \setminus A) \setminus B)$$. We want to prove $$x \in C \iff x \in D.$$
Assume $$x \in C$$. If $$x \in A \cup B$$, then $$x \in (A \cup B) \cap C$$ so $$x \in D$$.
If $$x \in A \cup B$$, then $$x \notin A$$ and $$x \notin B$$, so $$x \in C \setminus A$$ and also $$x \in (C \setminus A) \setminus B$$, so again $$x \in D$$.
Conversely, if $$x \notin C$$, then $$x$$ is not in either component of the union that comprises $$D$$ so $$x \notin D$$ and we are done.