Suppose that $(x,y)\in A\times(B\triangle C)$. Then $x\in A$ and $y\in B\triangle C$. The latter means that $y\in B$ or $y\in C$, but not both. In the first case, i.e., $y\in B$ but $y\notin C$, we have $(x,y)\in A\times B$. However, $(x,y)\notin A\times C$ because that would mean, in particular, that $y\in C$. Therefore, $(x,y)\in (A\times B)\triangle (A\times C)$. The second case ($y\in C$ but $y\notin B$) is considered similarly. This concludes the proof that $A\times(B\triangle C)\subseteq (A\times B)\triangle (A\times C)$. You can try proving the converse inclusion.daniel felipe said:AX(BΔC)=(AXB)Δ(AXC)