How might I solve this? I can't find any problem similar to this, and I always end up with the wrong terms.
If (AB) = 0 and (A+B) = 1, prove that (A+C)(!A+B)(B+C) = BC
Expanding the LHS is D=BC(A+!A+1)+!AC+AB=(B+!A)C+(AB). Since A+B=1 and AB=0, !A=B hence B+!A=B and D=(BC)+(AB). Once again, AB=0 hence D=BC.
did's answer is the way to go in general. However, notice that for your problem, $AB = 0$ and $A +B = 1$ implies that either $A = 1$ and $B = 0$ or $A = 0$ and $B = 1$.
In the first case, $BC = (B + !A) = 0$, and so the two sides are equal.
In the second case, we have
$$ (A +C)(!A + B)(B + C) = (C)(1)(1) = C = BC $$