Answer
$A\bigtriangleup B= (A - B) \cup (B - A) \\
=(B-A)\cup (A-B) \,\,\,\,\,\,\\(by\,\,commutative\,\,law\,\,for\,\,\cup)\\
but\,\,(B-A)\cup (A-B)=B\bigtriangleup A \\
so\,\,A\bigtriangleup B=B\bigtriangleup A
$
Work Step by Step
$A\bigtriangleup B= (A - B) \cup (B - A) \\
=(B-A)\cup (A-B) \,\,\,\,\,\,\\(by\,\,commutative\,\,law\,\,for\,\,\cup)\\
but\,\,(B-A)\cup (A-B)=B\bigtriangleup A \\
so\,\,A\bigtriangleup B=B\bigtriangleup A
$