Given that $A \cup B = A \cup C$, and $A \cap B = A\cap C$. Show that $B=C$. The problem:

You are given that $A \cup B = A \cup C$, and $A \cap B = A\cap C$. Show that $B=C$. 

My confusion: How does one approach such a problem? I know a few methods to solve such problems, but none of them work. I thought I could start by letting $x \in B$, and then deduce my way into proving that $x \in C$ as well, then proving the converse and finally showing that $B=C$. But I immediately see that this is a wrong approach, since I can't go anywhere from $x\in B$ with the given information. 
 A: Take $x\in B$ we want to show that $x\in C$.
Now let's see which property we can use, we can't use $A\cap B = A\cap C$ right now because $x$ is not necessarily an element in $A\cap B$. However we can use $A\cup B = A\cup C$, since $x\in B$ it is also an element in $A\cup B$ so $x\in A\cup C$.
If $x\in C$ we're done. Otherwise $x\in A$ but it's also in $B$ so $x\in A\cap B$. Now we can finally use the second property to get that $x\in A\cap C$ and so $x\in C$.
A: Take $b\in B$. Then, $b\in A\cup B=A\cup C$ so $b\in A$ or $b\in C$. If $b\in C$ you get the inclusion $B\subset C$. If $b\in A$, then $b\in A\cap B=A\cap C$ thus $b\in C$, also proving $b\in C$. The other inclusion works the same, interchanging roles of $B$ and $C$.
A: \begin{align}
B\cap A^C=(A\cup B)\cap A^C=(A\cup C)\cap A^C=C\cap A^C \\
B=(B\cap A)\cup(B\cap A^C)=(C\cap A)\cup(C\cap A^C)=C
\end{align}
A: Here is a simple proof using the basic properties of set operations:
$C = C\cup (A\cap C)$ by absorption.
This equals $C\cup (A\cap B)$ by hypothesis.
This equals $ (C\cup A)\cap (C\cup B)$ by distributivity.
This equals $(A\cup B)\cap (B\cup C)$ by hypothesis.
This equals $B\cup (A\cap C)$ by distributivity.
This equals $B\cup (A\cap B)$ by hypothesis.
This equals $B$ by absorption. 
Commutativity used here and there.
A: $B = B \cap (A \cup B) = B \cap (A \cup C) = (B \cap A) \cup (B \cap C) = (A \cap B) \cup (B \cap C) = (A \cap C) \cup (B \cup C) = (A \cup B) \cap C = (A \cup C) \cap C =C$
