Answer:
(I don't have the complement sign of sets, so I've used "©" to denote compliment)
L.H.S.
(B-A)∪C
= (B∩A©)∪C
= (B∪C)∩(A©∪C)
= (B∪C)∩(A-C)©
= (B∪C)-(A-C) = R.H.S.
(Proved)
