Answer
$for\,\,all\,\,sets\,\,A,B\,\,and\,\,C \\
(A\cap B)\cup (A\cap B^{c})=A \\
to\,\,prove\,\,this\,\,we\,must\,prove\,that\,\\
1-(A\cap B)\cup (A\cap B^{c})\subseteq A \\
2-A\subseteq (A\cap B)\cup (A\cap B^{c}) \\
proof\,of\,\,1:\\
x\in(A\cap B)\cup (A\cap B^{c}) \\
by\,\,def.\,\,of\,\,union\,\,of\,\,sets \\
x\in (A\cap B)\,or\,x\in (A\cap B^{c}) \\
and\,by\,\,def.\,\,of\,\,inter\! section \\
(x\in A\,and\,x\in B)\,or\,(x\in A\,and\,x\in B^{c})\\
in\,\,both\,cases\,x\in A \\
\because
x\in(A\cap B)\cup (A\cap B^{c}) \Rightarrow x\in A\\
\therefore (A\cap B)\cup (A\cap B^{c})\subseteq A \\$
$proof\,of\,\,2:\\
x\in A \,\Rightarrow x\in B\,\,or\,x\notin B \\
first\,case: (x\in A\,and\,x\in B)\\
\therefore x\in A\cap B \\
by\,\,def.\,\,of\,\,union \\
x\in (A\cap B)\cup (A\cap B^{c})\\
second\,case\,:(x\in A\,and\,x\notin B)\\
by\,\,def.\,\,of\,\,complement \\
x\in A\,and\,x\in B^{c} \\
\therefore x\in A\cap B^{c} \\
by\,\,def.\,\,of\,\,union \\
x\in (A\cap B^{c})\cup (A\cap B)\\
\because x\in A\Rightarrow x\in (A\cap B^{c})\cup (A\cap B)\\
(in\,both\,cases)\\
\therefore A\subseteq(A\cap B)\cup (A\cap B^{c})
$
$from\,\,1\,\,and\,\,2 \\
(A\cap B)\cup (A\cap B^{c})=A
$
Work Step by Step
$for\,\,all\,\,sets\,\,A,B\,\,and\,\,C \\
(A\cap B)\cup (A\cap B^{c})=A \\
to\,\,prove\,\,this\,\,we\,must\,prove\,that\,\\
1-(A\cap B)\cup (A\cap B^{c})\subseteq A \\
2-A\subseteq (A\cap B)\cup (A\cap B^{c}) \\
proof\,of\,\,1:\\
x\in(A\cap B)\cup (A\cap B^{c}) \\
by\,\,def.\,\,of\,\,union\,\,of\,\,sets \\
x\in (A\cap B)\,or\,x\in (A\cap B^{c}) \\
and\,by\,\,def.\,\,of\,\,inter\! section \\
(x\in A\,and\,x\in B)\,or\,(x\in A\,and\,x\in B^{c})\\
in\,\,both\,cases\,x\in A \\
\because
x\in(A\cap B)\cup (A\cap B^{c}) \Rightarrow x\in A\\
\therefore (A\cap B)\cup (A\cap B^{c})\subseteq A \\$
$proof\,of\,\,2:\\
x\in A \,\Rightarrow x\in B\,\,or\,x\notin B \\
first\,case: (x\in A\,and\,x\in B)\\
\therefore x\in A\cap B \\
by\,\,def.\,\,of\,\,union \\
x\in (A\cap B)\cup (A\cap B^{c})\\
second\,case\,:(x\in A\,and\,x\notin B)\\
by\,\,def.\,\,of\,\,complement \\
x\in A\,and\,x\in B^{c} \\
\therefore x\in A\cap B^{c} \\
by\,\,def.\,\,of\,\,union \\
x\in (A\cap B^{c})\cup (A\cap B)\\
\because x\in A\Rightarrow x\in (A\cap B^{c})\cup (A\cap B)\\
(in\,both\,cases)\\
\therefore A\subseteq(A\cap B)\cup (A\cap B^{c})
$
$from\,\,1\,\,and\,\,2 \\
(A\cap B)\cup (A\cap B^{c})=A
$