Is it possible to simplify the boolean expression $A'.B + A.B'$?

Question

I know it's just an easy question but I can't get it at all.

A'.B + A.B'

Is it possible to simplify this expression?

Actually, I got this expression from this

A'.B.C+A.B'.C+A.B.C'+A.B.C = F
C(A'.B+A.B') + A.B(C'+C) = F
C(A'.B+A.B') + A.B = F

Now, I'm stuck at here.

Answer 1

Finally, I figured it out. I would like to share the answer here for those wondering.

A'.B.C + A.B'.C + A.B.C' + A.B.C
A'.B.C + A.B'.C + A.B(C' + C)
A'.B.C + A.B'.C + A.B
A'.B.C + A(B'.C + B)
A'.B.C + A(B + C) [Distributive law]
A'.B.C + A.B + A.C
B(A'.C + A) + A.C
B(A + C) + A.C

A.B + B.C + A.C = F //