I don't know exactly how to simply this problem. I can clearly see that (A + B) is in all of them but I don't know what to do next.
(A + B + C)(A + B + !C + D)(A + B + !C + !D)
-- Edit 1 --
I am trying to get something like (A + B) + [ (C)(!C + D)(!C + !D) ]. I know this is wrong but this is where I'm trying to get to as the next step.
-- Edit 2 --
The following steps are from Edit one.
(A + B) + [ (C)(!C + D)(!C + !D) ]
= (A + B) + [ (C!C + CD)(!C + !D) ]
= (A + B) + [ (0 + CD)(!C + !D) ]
= (A + B) + [ (CD)(!C + !D) ]
= (A + B) + [ (CD!C) + CD!D) ]
= (A + B) + [ (0) + 0) ]
= (A + B)
-- Edit 3 --
I verified using a Boolean calculator and it seems that the answer is correct.
Thank you.
Here is a handy equivalence:
Adjacency
$(P + Q)(P+ !Q)=P$
With that, we can simplify your expression in exactly two steps:
$$(A + B + C)(A + B + !C + D)(A + B + !C + !D)\overset{Adjacency}=$$
$$(A + B + C)(A + B + !C)\overset{Adjacency}=$$
$$A + B$$
Yeah, a good one to put in your boolean algebra toolbox!