Simplifying with boolean algebra.

Question

I have two parts to my question, the first being how do I use Boolean algebra to reduce A'C'+ABC+AC' to have only 3 literals. I sort of understood the concept where you can factor things out but for some reason it won't come to mind with this one. My next question is how do I use DeMorgan's law with (a+c)(a+b')(a'+b+c').

I don't necessarily need the answers, I need to know the process to get the solution.

Edit: For DeMorgan's law, I'm not really sure if this is right but I multiplied out all 3 and got aa'+aa'b'+aa'c+a'b'c+ab+abb"+abc+bb'c+ac'+ab'c'+acc'+b'cc'.

If that is right then the next step would be to simplify correct?

Answer 1

HINT

For the first problem focus on the terms $A'C'$ and $AC'$

You can do: $A'C' + AC' = (A' + A)C' = (1)C' = C'$

So as the first few steps you get:

$A'C' + ABC + AC' = $

$A'C' + AC' + ABC = $

$(A' + A)C' + ABC = $

$(1)C' + ABC = $

$C' + ABC$

Now, do you see how you can simplify this just a little bit more?