Simplify Boolean Algebra How do I simplify the following expression with Boolean Algebra? Please show what you used to simplify so I can understand.

$$ABC + AB'C' + ABC' + A'B'C'$$

 A: Note that you have both $ABC$ and $ABC'$: 
$$ABC+ABC'=AB(C+C')=AB\;.$$
The rest is $AB'C'+A'B'C'$, and you can use the same idea:
$$AB'C'+A'B'C'=(A+A')B'C'=B'C'\;.$$
After those simplifications the expression has been reduced to $AB+B'C'$, which is as simple as it’s going to get.
A: First I want to group the elements that are similar. This will allow me to start reducing the expression.
$$ABC + AB'C' + ABC' + A'B'C'$$
$$ABC + ABC' + AB'C' + A'B'C'$$
$$[AB(C+C')] + [B'C'(A+A')] // Group.$$
$$AB + B'C'                 // α+α' = 1; α * 1 = α $$    
Edit: For me Boolean Algebra is superior, but if you ever get stuck on a problem try using a Karnaugh map: http://en.wikipedia.org/wiki/Karnaugh_map
If you read the Wiki page you will see that with a Karnaugh map you can simplify this expression.
A: Remember that $C+C'=1$ and $X\cdot1=X$, and $X(Y+Z)=XY+XZ$. Then we have:
$$\begin{align}
&ABC+AB'C'+ABC'+A'B'C'=\\
&ABC+ABC'+AB'C'+A'B'C'=\\
&AB(C+C')+(A+A')B'C'=\\
&&=AB+B'C'
\end{align}$$
A: First of all we will rearrange a question to make it easy.
ABC+AB′C′+ABC′+A′B′C′
                                                    boolean laws.

Solution:
                                                     (C+C')=1
      1)STEP:  ABC+ABC'+AB'C'+A'B'C'

      2)STEP:  AB(C+C')+(A+A')B'C'                  (A+A')=1  
      3)STEP:  AB(1)+(1)B'C'
 RESULT ANSWER:
              AB+B'C'

Can find boolean laws truth table on this link
[1]: https://www.electronics-tutorials.ws/boolean/bool_6.html
