How to Disprove $A-(B-C)=A-(B\cup C)$? For $A,B,C$ sets. I know I want to show that it is not the case that $A-(B-C)$ is not equal to $A-(B\cup C)$, I also know that the definition of set deference is x is an element in A but not in B if I have set A-B, but to disprove this is getting a bit difficult. 

While the other answers already gave counterexamples, and while truth tables will always point you to sufficient counterexamples, the following is more about how to "look for" easy counterexamples.

I want to show that it is not the case that $A-(B-C)$ is not equal to $A-(B\cup C)$

The difference between the two expressions is $\;B \setminus C\;$ vs. $\;B \cup C\;$. The former is a (smaller) subset of $B$ since $\;B \setminus C \subseteq C\;$, while the latter is a (larger) superset of $B$ since $B \cup C \supseteq B\,$. When does this difference get exacerbated the most? When $B$ is the "smallest" possible, so consider $B = \emptyset\,$.
For $B=\emptyset$ obviously $\;B \setminus C = \emptyset\;$ and $B \cup C = C\,$, so $A \setminus (B \setminus C) = A$ and $A \setminus (B \cup C)=A \setminus C$. Again, look at the difference between the two sides, it's the $\setminus C$ on the RHS. When does that difference get exacerbated the most? When $C$ is the "largest" possible, and since we are only interested in the relation to $A$, that means $C=A$.
So in the end, one natural counterexample is $B=\emptyset, C=A$ for which:
$$A \setminus (B \setminus C) = A \setminus (\emptyset \setminus A) = A \setminus \emptyset = \;A \quad\ne\quad \emptyset\; = A \setminus A = A \setminus(\emptyset \cup A) = A \setminus (B \cup C)
$$
(Note: the latter holds true for $A = \emptyset$ of course, but the question was about arbitrary sets, so this still is a sufficient counterexample.)
