I'll write A' for , and similarly for other variables.

. So, by using distributivity and eliminating the same minterms, or disjuncts (AB and twice ABC'), we get

AB + AC + ABC' + BC + AC'

Next, ABC' + AC' = AC'(B + 1) = AC', so we have

AB + AC + BC + AC'

AC + AC' = A, and A + AB = A, so the result is A + BC.