Can't simplify this boolean expression I'm trying to simplify this boolean expression:
$$(AB)+(A'C)+(BC)$$
I'm told by every calculator online that this would be logically equivalent:
$(AB)+(A'C)$
But so far, following the rules of boolean algebra, the best that I could get to was this: 
$(B+A')(B+C)(A+C)$
All of the above are logically equivalent (I've made a truth table for each) but I don't understand what steps am I missing trying to simplify the expression.
I also couldn't find an "expression simplifier" tool online that could show me the steps that I'm missing.
Help / directions to go to would be much appreciated, thanks in advance.
 A: Well, clearly there's either $A$ or $A'$ in the first two terms, so use this to split the third wheel up, and absorb the pieces.
$\begin{align}(AB)+(A'C)+(BC) & = (AB)+(A'C)+(A+A')(BC) \\ & = (AB)+(A'C)+(ABC)+(A'BC) \\ & = (AB+ABC)+(A'C+A'BC) \\ & = (AB)(1+C)+(A'C)(1+B) \\ & = (AB)+(A'C)\end{align}$
A: I would like to add the following explanation to the above answers:

The first two terms translate as “If A, then B, else C ”. Notice, therefore, that B and C cannot  simultaneously coexist, meaning that the third term can be safely ignored or omitted, since it  is superfluous or redundant.

A: This is known as the consensus rule.
$BC$ is the "consensus" term of $AB$ and $A'C$ and can be removed or added to the boolean expression.
Derivation
$$(X.Y)+(X'.Z)+(Y.Z)$$
$$=(X.Y)+(X'.Z)+(X+X')(Y.Z)$$
$$=(X.Y)+(X'.Z)+(X.Y.Z)+ (X'.Y.Z)$$
$$=(X.Y)+(X.Y.Z)+(X'.Z)+(X'.Y.Z)$$
$$=X.Y(1+Z) + X'.Z(1+Y)$$
$$=X.Y + X'.Z$$
Thus $$(X.Y)+(X'.Z)+(Y.Z)= X.Y + X'.Z$$
A: $$\begin{align*}
AB+A'C+BC&=AB(C+C')+A'C(B+B')+BC\\
&=ABC+ABC'+A'BC+A'B'C+BC\\
&=(A+A')BC+BC+ABC'+A'B'C\\
&=BC+BC+ABC'+A'B'C\\
&=BC+ABC'+A'B'C\\
&=(A+A')BC+ABC'+A'B'C\\
&=ABC+A'BC+ABC'+A'B'C\\
&=AB(C+C')+A'(B+B')C\\
&=AB+A'C
\end{align*}$$
A: Though this has already some good answer, I know an interesting way to look at boolean expression with less than 4 elements(for greater than 3 it becomes messy ),it might be helpful for any future user struggling with a similar problem.
Imagine them as Venn diagrams, draw their intersection using the given expression, after this most of the time question becomes trivial.
You can easily see/confirm A'C +AB covers the colored region completely.
(ignore my drawing skills)

