Prove that $A \times (B\oplus C) = (A \times B) \oplus (A \times C)$ $A \times (B\oplus C) = (A \times B) \oplus (A \times C)$
$LHS \Rightarrow RHS$
$A \times [(B\land\neg C)\lor (\neg B \land C)]$
{$(a,b): a \in A, b \in (B\land\neg C)\lor (\neg B \land C)$}
and then im stuck here, not sure what else to do with the cartesian product
 A: From where you left off, you have $a \in A,b\in B,b\notin C \vee a\in A, b \notin B, b \in C$.
Hence, $(a,b) \in (A \times B) \vee (A \times C)$ and  $b\in B \Leftrightarrow b\notin C$ so that $(a,b)\in (A \times B) \Leftrightarrow (a,b) \notin (A \times C)$.
Therefore $(a,b) \in (A \times B) \oplus (A \times C)$ and $A \times (B\oplus C) \subseteq (A \times B) \oplus (A \times C).$
The other implication is proven similarly.
A: Just note the notation (even you might know what you are talking about) is not correct

$$A \times [(B\land\neg C)\lor (\neg B \land C)]$$

Is this a set or a proposition? If it's a set
$$A \times (B\oplus C)=A\times[(B\cap C^c)\cup(B^c\cap C)]$$
If it's a proposition, for $(a,b)\in A \times (B\oplus C)$ have
$$a\in A\land b\in[(B\cap C^c)\cup(B^c\cap C)]$$
To show this equation hold, from definition we can write
\begin{align}
(A\times B)\oplus(A\times C)=&\{(a,b):(a,b)\in(A\times B)\land (a,b)\not\in(A\times C)\\
&\lor(a,b)\not\in(A\times B)\land(a,b)\in(A\times C)\}\\
=&\{(a,b):(a\in A\land b\in B)\land(a\in A\to b\notin C)\\
&\lor(a\in A\to b\not\in B)\land(a\in A\land b\in C)\}\\
=&\{(a,b):(a\in A\land b\in B\land b\not\in C)\\
&\lor (b\not\in B\land a\in A\land b\in C)\}\\
=&\{(a,b):a\in A\land [(b\in B\land b\not\in C)\lor (b\not\in B\land b\in C)]\}\\
=&\{(a,b):a\in A\land (b\in B\oplus C)\}\\
=&A \times (B\oplus C)
\end{align}
