Let $A$, $B$, and $C$ are three events The question is: 
$A$, $B$ and $C$ are three events, subset of $\Omega$, then
what should be the condition to have 
$(A\triangle B)\cup(A\triangle C)=A\triangle(B\cup C)$.
Then here was my attempt;
We have $(A\triangle B)\cup(A\triangle C)=[(A\cap B^c)\cup(A^c\cap B)]\cup[(A\cap C^c)\cup(A^c\cap C)]$
$=(A\cap(B^c\cup C^c))\cup(A^c\cap(B\cup C))$
$=A\triangle (B\cup C)$
Then, $(B\cup C)^c$ = $(B^c\cup C^c)$. But we know by De Morgan's Law, $(B\cup C)^c=B^c\cap C^c$, right? So are $B$ and $C$ equal? Why?
 A: The condition you require is $$A\cap B=A\cap C.$$
By far the easiest way to prove this is to draw a Venn diagram for the given equation.
To prove it algebraically, note that you have correctly put the LHS into the form:- 
$$(A\cap(B^c\cup C^c))\cup(A^c\cap(B\cup C)).$$
The RHS is 
$$(A\cap(B^c\cap C^c))\cup(A^c\cap(B\cup C)).$$
These two expressions will be equal if their intersections with $A$ and $A^c$ are both equal and therefore we only need consider 
$$A\cap(B^c\cup C^c)=A\cap(B^c\cap C^c)\;\; *$$
Intersecting with $C$ now gives
$A\cap B^c\cap C=0$ i.e $A\cap C$ is contained in $B$. Similarly $A\cap B$ is contained in $C$ and so $A\cap B=A\cap C.$
Conversely, equation * clearly holds when $A\cap B=A\cap C$. 
A: 
The question is:  $A$, $B$ and $C$ are three events, subset of
  $\Omega$, then what should be the condition to have 
$(A\triangle B)\cup(A\triangle C)=A\triangle(B\cup C)$.

If you assume $C\subseteq B$ and $A\cap B =\emptyset$,
\begin{align}
(A\triangle B)\cup(A\triangle C) & = \left((A\backslash B)\cup(B\backslash A)\right)\cup\left((A\backslash C)\cup(C\backslash A)\right)  \\ & = 
(A\backslash B)\cup(B\backslash A)\cup(A\backslash C)\cup(C\backslash A) \\ & = (A\backslash(B\cup C))\cup((B\cup C)\backslash A)\cup A \cup C  \\ & =
A\triangle(B\cup C) \cup A \cup C \\ & =
A\triangle(B\cup C) 
\end{align}
This is most likely not the best solution, but it works nonetheless.
