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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??

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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:

$A'BC + AB'C + ABC' + ABC = \text{ (Idempotence x 2)}$

$A'BC + ABC + AB'C + ABC + ABC' + ABC = \text{ (Adjacency x 3)}$

$BC + AC + AB$

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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}

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