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The Boolean expression for the following would be:
$F(A,B,C) = \Sigma(0,2,6)$
$d(A,B,C) = \Sigma(1,3,5)$
Minterms:
$0 = 000$
$2 = 010$
$6 = 110$
Don't care terms:
$1 = 001$
$3 = 011$
$5 = 101$
Using K-map simplification:
Group $(0,2)$ → $A'C'$
Group $(2,6)$ → $BC'$
Thus,
$F = A'C' + BC'$
Factorizing:
$F = C'(A' + B)$
Which simplifies to:
$F = A' + BC'$
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