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- $A\ast A = AA+A'A' = A + A' = 1$
- $C\ast A= (A\ast B)\ast A = (AB + A'B')\ast A= (AB + A'B')A +(AB + A'B')'A'$
$\qquad =(AB + A'B')A+(A'B+AB')A' = AB +0 +A'B+0 = B.$
- $AB + A'C= 1, AC + B = 0$
$\implies AC + B = 0,$ means both $B = 0$ and $AC = 0$
$\implies AB+ A'C = 1$
$\implies A'C = 1 \qquad \left[\because B = 0, AB = 0 \right]$
So, $C = 1$ and $A = 0$
$A = 0 , B = 0$ and $C = 1$