Another way is to note that $\displaystyle B = (A\vartriangle B)\vartriangle A$, and similarly for $\displaystyle C$. This in turn can be proved by showing that $\displaystyle x\in (A\vartriangle B) = (x\in A)\oplus (x\in B)$ where $\displaystyle x\in X=1$ or $\displaystyle 0$ depending on whether $\displaystyle x$ is in $\displaystyle X$ and $\displaystyle \oplus$ is exclusive or. A similar method for proving associativity of $\displaystyle \vartriangle$ is described in

this post. If one starts with knowing that $\displaystyle \vartriangle$ is associative and commutative then $\displaystyle B = (A\vartriangle B)\vartriangle A$ is obvious.