Answer
$A\bigtriangleup A = (A - A) \cup (A - A) \\
but(A-A=\varnothing \,\,\,by\,\,def.\,\,of\,\,set\,\,dif\! ference\,\,)\\
A\bigtriangleup A = (A - A) \cup (A - A) =\varnothing \cup \varnothing =\varnothing \\
\therefore A\bigtriangleup A =\varnothing
$
Work Step by Step
$A\bigtriangleup A = (A - A) \cup (A - A) \\
but(A-A=\varnothing \,\,\,by\,\,def.\,\,of\,\,set\,\,dif\! ference\,\,)\\
A\bigtriangleup A = (A - A) \cup (A - A) =\varnothing \cup \varnothing =\varnothing \\
\therefore A\bigtriangleup A =\varnothing
$