I'm trying to simplify the following boolean expression:
[(A' (C+D)')'] (A) + ( B (DC) + (D'C') + A + CB'
What I got is
A + (C+D) + B [(DC) + (D'C')] + A + CB'
A(A+C) + D + B[1] + A +CD'
A + D + B + A +CD' (used absoprtion law for A(A+C)
A + B + D + CD'
are my steps correct? if not, where did I get it wrong??
Your first step isn’t quite right:
$$\big(A'(C+D)'\big)'A=AA+(C+D)A=A+(C+D)A=A+AC+AD\;;$$
you forgot the factor of $A$ on the $(C+D)$ term. Note that this can be further simplified to $A$ by absorption.
There are several problems with your second step. One seems to be a typo: the $CB'$ at the end should be $CD'$. More important, $A+C$ is not equivalent to $A(A+C)$, and $DC+D'C'$ is not equivalent to $1$.
Your initial expression has an unmatched parenthesis: $(B(DC)+(D'C')$. What you have after your first step suggests that this was supposed to be $B[(DC)+(D'C')]$. If so, it should expand (with the simplification noted above) to
$$A+BCD+BC'D'+A+B'C\;,$$
which immediately simplifies to
$$A+BCD+BC'D'+B'C\;.$$
This can be rewritten in various ways (e.g., $A+CD+B'CD'+BC'D'$), but I don’t immediately see anything substantially simpler.
Note that your final answer cannot be right: it’s $1$ when $A=1$, no matter what $B,C$, and $D$ are, but you can find values for $B,C$, and $D$ that make the original expression $0$ when $A=1$.