To simplify the Boolean expression using a Karnaugh map (K-map), we will first translate the given Boolean expression.

The expression is:

$A = \overline{X}\overline{Y}Z + \overline{X}Y\overline{Z} + X\overline{Y}\overline{Z} + X\overline{Y}Z + XY\overline{Z} + \overline{X}YZ$

This involves three variables $X$, $Y$, and $Z$. Let me explain the steps:

1. List the Minterms: Each term in the Boolean expression corresponds to a minterm. Here are the minterms from the expression:

   • $\overline{X}\overline{Y}Z$ corresponds to minterm $001$ (decimal 1).
   • $\overline{X}Y\overline{Z}$ corresponds to minterm $010$ (decimal 2).
   • $X\overline{Y}\overline{Z}$ corresponds to minterm $100$ (decimal 4).
   • $X\overline{Y}Z$ corresponds to minterm $101$ (decimal 5).
   • $XY\overline{Z}$ corresponds to minterm $110$ (decimal 6).
   • $\overline{X}YZ$ corresponds to minterm $011$ (decimal 3).

2. Create the K-map: We'll fill in the K-map with 1's for these minterms (1, 2, 3, 4, 5, and 6).

   	Y'Z'	Y'Z	YZ	YZ'
   X'Y'	0	1	0	0
   X'Y	1	0	1	0
   XY'	1	1	0	1
   XY	0	0	0	0

3. Group the 1's:

   • We can group the 1's in pairs or groups of four if possible.
   • Group the minterms 1, 5, and 4, 6, as well as the minterms 2 and 3.

4. Simplified Expression: The simplified Boolean expression after grouping will be:

   $A = \overline{Y}Z + X\overline{Z}$

Thus, the simplified Boolean expression is $A = \overline{Y}Z + X\overline{Z}$.

Would you like more details on K-map grouping, or have any other questions?

Here are 5 follow-up questions related to the topic:

1. How does grouping work in a K-map for more than two variables?
2. Can a Boolean expression be simplified further after a K-map?
3. What are prime implicants in Boolean algebra?
4. How does De Morgan’s Law apply in Boolean simplification?
5. How can don’t-care conditions be handled in a K-map?

Tip: In a K-map, always aim to group the largest number of 1's possible to achieve the simplest expression.