understanding Boolean algebra ( Boolean algebra simplification ) F = A'BC + AB'C + ABC' + ABC

I know that by K-map answer is AB+AC+BC, and by boolean rules, it will be
=A'BC + AB'C + ABC' + ABC+ ABC + ABC 
=(A'BC + ABC) + (AB'C + ABC) + (ABC' + ABC)
=(A' + A) BC + (B' + B) CA + (C' + C) AB
=AB+AC+BC 

I can't get it, how we knew that we should add ABC to it??
 A: If you look at the K-Map, you'll find that as you create groupings, you use the $ABC$ cell three times. 
Using algebra we can do the same combinations, but that does mean we need $3$ $ABC$ terms.
And by the way, there is a handy boolean algebra rule that reflects the combining of adjacent cells:
Adjacency
$PQ + PQ' = P$
Applied to your our expression:
$A'BC + AB'C + ABC' + ABC = \text{ (Idempotence x 2)}$
$A'BC + ABC + AB'C + ABC + ABC' + ABC = \text{ (Adjacency x 3)}$
$BC + AC + AB$
A: You can observe that you can pair $BC$ in the first item with $BC$ in the fourth one:
$$
F=(A'+A)BC+AB'C+ABC'=BC+AB'C+ABC'
$$
However, you could also pair $AC$ in the second item with $AC$ in the fourth one:
$$
F=A'BC+(B'+B)AC+ABC'=AC+A'BC+ABC'
$$
Similarly, pairing $AB$ in the third item with $AB$ in the fourth one:
$$
F=A'BC+AB'C+(C'+C)AB=A'BC+AB'C+AB
$$
Note that none of these final expressions is symmetric in $A$, $B$ and $C$, whereas the original one is.
This gives the idea of doing the three pairings all together, which is obtained by adding $ABC$ twice, justified because $P+P=P$:
\begin{align}
F
&=A'BC+AB'C+ABC'+ABC \\
&=A'BC+ABC+AB'C+ABC+ABC'+ABC\\
&=(A'+A)BC+(B'+B)AC+(C'+C)AB\\
&=BC+AC+AB
\end{align}
Alternatively, we get a symmetric expression by summing up the three we got:
\begin{align}
F
&=F+F+F\\
&=(BC+AB'C+ABC')+(AC+A'BC+ABC')+(A'BC+AB'C+AB)\\
&=AB+ABC'+BC+A'BC+AC+AB'C\\
&=AB(1+C')+BC(1+A')+AC(1+B')\\
&=AB+BC+AC
\end{align}
