How to simplify circuits

Question

The answer to the question is A in case someone thinks Im cheating for a test or something. I really wan't to understand how to solve such questions. I am guessing there is some way to bring these brackets together and to get a simplified circuit but I don't understand it fully. Maybe something regarding truth tables?

Answer 1

  abc' + a'bc' + a'b'c' + ab'c'        initial expression
= (a + a')bc' + a'b'c' + ab'c'         distributive law
= (T)bc' + a'b'c' + ab'c'              additive inverse
= bc' + a'b'c' + ab'c'                 multiplicative identity
= bc' + (a' + a)b'c'                   distributive law
= bc' + (T)b'c'                        additive inverse
= bc' + b'c'                           multiplicative identity
= (b + b')c'                           distributive law
= (T)c'                                additive inverse
= c'                                   multiplicative identity

The expression is equivalent to the inverse of c and so one logic gate is enough.