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I've figured out the following statement is true but I was wondering how you actually go about proving something like this?

$A \times (B \triangle C) = (A \times B) \triangle (A \times C)$

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    One possible approach: to show that two sets are equal, prove two inclusions, i.e. that $\text{LHS}\subseteq\text{RHS}$ and that $\text{RHS}\subseteq\text{LHS}$. And each inclusion proof can be done element-wise: for example, pick an arbitrary $x\in\text{LHS}$ and by using the definitions of these set operations show that $x\in\text{RHS}$. – 2017-02-15

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from my answer in question:

Def. let be $E,F$ sets: $$E \bigtriangleup F := (E \setminus F) \cup (F\setminus E)$$

The. let be $A,B,C$ sets: $$A \times (B \bigtriangleup C)=(A \times B) \bigtriangleup (A \times C)$$ Proof: $$ \begin{align} A \times (B \bigtriangleup C) &= A \times ((B \setminus C) \cup (C \setminus B)) \\&= (A \times (B\setminus C)) \cup (A \times (C \setminus B)) \\&=((A \times B)\setminus (A \times C)) \cup ((A \times C) \setminus (A \times B)) \\&=(A\times B)\bigtriangleup (A \times C)\end{align}$$ I used "Cartesian Product Distributes over Union" and "Cartesian Product Distributes over Set Difference"

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A nice approach is by comparing the characteristic functions. Note that :

$(x,y)\in A\times B\Leftrightarrow \left(x\in A\,\mathrm{and}\,y\in B\right)$

so that $1_{A\times B}=1_A\times 1_B$

Also $1_{A\Delta B}=1_A\oplus A_B$ where $\oplus$ denotes mod 2 addition.

Now the required proof is straightforward.