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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

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I've removed algebra tag, since we don't use algebra tag anymore, see meta for details. – Martin Sleziak Oct 22 '12 at 8:58
up vote 1 down vote accepted

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.

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Can you explain "Since A+B=1, B+!A=B"? What is the logic in this statement? – user1764389 Oct 22 '12 at 6:21
@user1764389: $A + B = 1$ and $AB = 0$ implies that $A = !B$. Hence, $B + !A = B + B = B$. – JavaMan Oct 22 '12 at 7:25

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

$$ (A +C)(!A + B)(B + C) = (C)(1)(1) = C = BC $$

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