Finding Product-of-Maxterms Form I need help to resolve this problem, i have the following boolean function:
[(A.!C)+!(A.!C)].!(A.!B)

The Truth table is:

(please see this LINK TO wolframalpha for more detail)
Then the Sum-of-Minterms Form is:
A.B.C + A.B.!C + !A.B.C + !A.B.!C + !A.!B.C + !A.!B.!C

And the Product-of-Maxterms Form is:
(A + !B + C).(A + !B + !C) 

¿Is this correct? ¿How do I demonstrate this? Because I tried to find the maxterms starting from the initial Boolean function, for example:
 [(A.!C)+!(A.!C)].!(A.!B)
= [(A.!C)+!(A.!C)].!(A.!B)
= (A.!C + !A + C).(!A + B)
= (!A + !C + C).(!A + B)
= (!A + 1).(!A + B)
= !A + B
= .....
= (A + !B + C).(A + !B + !C)  ???????????????

I do not understand how to find (A + !B + C).(A + !B + !C).
Please appreciate any help. Thanks.
 A: Let's call your expression $f$,
$$
\DeclareMathOperator{\and}{~And~}
\DeclareMathOperator{\or}{~Or~}
\DeclareMathOperator{\nt}{~Not~}
\DeclareMathOperator{\T}{{\color{blue}T}}
\DeclareMathOperator{\F}{{\color{red}F}}
f = \bigg((A \and \nt C) \or \nt (A \and \nt C)\bigg) \and \nt (A \and \nt B) $$
$$\begin{array} {c|ccc|c}
& A & B & C & f \\ \hline
\text{Row 1} & \T & \T & \T & \T \\
\text{Row 2} & \T & \T & \F & \T \\
\text{Row 3} & \T & \F & \T & \F \\
\text{Row 4} & \T & \F & \F & \F \\
\text{Row 5} & \F & \T & \T & \T \\
\text{Row 6} & \F & \T & \F & \T \\
\text{Row 7} & \F & \F & \T & \T \\
\text{Row 8} & \F & \F & \F & \T \\
\end{array}$$
Notice how your sum of minterms has 6 terms.  Your truth table also has 6 "true" entries.  To construct minterms, you just list the "true" cases:
$$\begin{align}f &= 
(\text{Row 1}) \or 
(\text{Row 2}) \or 
(\text{Row 5}) \or 
(\text{Row 6}) \or 
(\text{Row 7}) \or 
(\text{Row 8})\\
&= ABC \or AB\overline C \or \overline ABC \or \overline AB\overline C \or \overline A\overline BC \or \overline A\overline B\overline C\end{align}$$
To construct the product of maxterms, look at the value of $\nt f$.  There are 2 cases when $f$ is false:
$$\begin{align}\nt f &= (\text{Row 3}) \or (\text{Row 4}) \\
&= A\overline B C \or A \overline B \overline C\end{align}$$
Now apply DeMorgan's:
$$\begin{align}
\nt\nt f &= \nt(A\overline B C \or A \overline B \overline C) \\
       f &= (\overline A\or B\or \overline C) \and (\overline A \or B \or C)
\end{align}$$
Your given answer for "product of maxterms" is the value of $f(\nt A, \nt B, \nt C)$.  Perhaps you are working with a different definition.  
