Is it true that $A\cup(B\mathbin{\triangle}C)=(A\cup B)\mathbin{\triangle}(A\cup C)$? Prove or disprove: 

$$A\cup(B\mathbin{\triangle}C)=(A\cup B)\mathbin{\triangle}(A\cup C)$$ for all sets $A,B,C$.

I'm confused as to how to do the steps here logically.
 A: The first step is to figure out if the damned thing is true or false. We can do that with a case analysis. A given element is either in $A$ or not in $A$, in $B$ or not in $B$, in $C$ or not in $C$, a total of $8$ cases ($2\times2\times2$). Let's start by supposing that $x$ is in $A$. 
Wow, in this case $B$ and $C$ are irrelevant! If $x\in A$, then $x$ is an element of the left side, regardless of $B$ and $C$; on the other hand, $x$ is not an element of the right side, again regardless of $B$ and $C$. That means the equation is false if there is anything at all in $A$, i.e., if $A$ is nonempty. We might now observe that the equation is true if and only if $A=\emptyset$ but that would be doing more that was required; we were only asked to prove or disprove the identity, not to find all cases where equality holds. All we have to do for a disproof is give one counterexample. Like this:
Let $A=B=C=\{3\}$. Then$$A\cup(B\mathbin{\triangle}C)=\{3\}\cup\emptyset=\{3\}$$and$$(A\cup B)\mathbin{\triangle}(A\cup C)=\{3\}\mathbin{\triangle}\{3\}=\emptyset$$so $A\cup(B\mathbin{\triangle}C)\ne(A\cup B)\mathbin{\triangle}(A\cup C)$.
If I was feeling lucky, I might have just started off by picking three random sets for $A,B$ and $C$ and hoping for a counterexample. And that would have worked, unless I was so unlucky as to pick $A=\emptyset$. Note that testing the equality with something like $$A=\{1,3,5,7\},B=\{2,3,6,7\},C=\{4,5,6,7\}$$ is tantamount to drawing the Venn diagram.
A: For elementary set-theoretic equalities like this the standard approach is to show that each side is a subset of the other. One often does that by ‘element-chasing’: to show that 
$$A\cup(B\mathbin{\triangle}C)\subseteq(A\cup B)\mathbin{\triangle}(A\cup C)\;,$$
for instance, let $x$ be an arbitrary element of $A\cup(B\mathbin{\triangle}C)$, and try to show from this that $x\in(A\cup B)\mathbin{\triangle}(A\cup C)$. I’ll do this half and leave the other half to you.
Suppose that $x\in A\cup(B\mathbin{\triangle}B)$. Now use the definitions of the various set operations. By the definition of union we know that $x\in A$ or $x\in B\mathbin{\triangle}C$. Suppose that $x\in A$; does this guarantee that $x\in(A\cup B)\mathbin{\triangle}(A\cup C)$? No: in fact, it guarantees that $x$ is not in $(A\cup B)\mathbin{\triangle}(A\cup C)$. To see this, use the definition of symmetric difference:
$$(A\cup B)\mathbin{\triangle}(A\cup C)=\Big((A\cup B)\setminus(A\cup C)\Big)\cup\Big((A\cup C)\setminus(A\cup B)\Big)\;.$$
Since $x\in A$, we know that $x\in A\cup B$ and $x\in A\cup C$, so $x\notin(A\cup B)\setminus(A\cup C)$ and similarly, $x\notin(A\cup C)\setminus(A\cup B)$, so $x\notin(A\cup B)\mathbin{\triangle}(A\cup C)$. This shows that if $A\ne\varnothing$, then 
$$A\cup(B\mathbin{\triangle}C)\nsubseteq(A\cup B)\mathbin{\triangle}(A\cup C)\;,$$
and therefore
$$A\cup(B\mathbin{\triangle}C)\subseteq(A\cup B)\mathbin{\triangle}(A\cup C)\;.$$
In other words, we’ve actually proved that the equality does not necessarily hold. For a concrete example, let $A=\{1\}$ and $B=C=\varnothing$: then
$$A\cup(B\mathbin{\triangle}C)=\{1\}\cup(\varnothing\mathbin{\triangle}\varnothing)=\{1\}\cup\varnothing=\{1\}\;,$$
but
$$(A\cup B)\mathbin{\triangle}(A\cup C)=\big(\{1\}\cup\varnothing\big)\mathbin{\triangle}\big(\{1\}\cup\varnothing\big)=\{1\}\mathbin{\triangle}\{1\}=\varnothing\;.$$
Added: If you make a Venn diagram first, you can save yourself most of this trouble: the regions corresponding to $A\cup(B\mathbin{\triangle}C)$ and $(A\cup B)\mathbin{\triangle}(A\cup C)$ are clearly different, so you need only produce a concrete example for which the equality fails. I could have said this at the beginning, but your comment about not knowing how to do the steps logically suggested that it might be useful also to illustrate what you would have to do to prove something that was true.
A: When I saw the question my intuition was this: $A$ is a subset of $A \cup (B \mathbin{\triangle} C)$, but since it appears in both sides of the $\mathbin{\triangle}$ in $(A \cup B) \mathbin{\triangle} (A \cup C)$ its elements are likely to disappear. Why? Because $X \triangle Y$ contains everything that lies in either $X$ or $Y$, but not both: anything that appears twice is thrown away. 
So consider $A=\{1 \}$, $B=\{2\}$ and $C=\{3\}$. Then
$$A \cup (B \mathbin{\triangle} C) = \{1\} \cup (\{ 2 \} \cup \{3\}) = \{ 1 \} \cup \{2,3\} = \{1,2,3 \}$$
but
$$(A \cup B) \mathbin{\triangle} (A \cup C) = \{1,2\} \mathbin{\triangle} \{1,3\} = \{2,3\}$$
These sets are not equal, so the equation does not hold.
