Can someone help me simplify this boolean algebra

Question

ABCD + AB(CD)' + (AB)'CD when i used basic rule it becomes weird but boolean calculator shows something else

The Question is to simplify the expression using the boolean algebra so My solution was ABCD + AB(CD)' + (AB)'CD =AB(CD + (C'+D') + (AB)'CD =AB(1) + (A'+B')CD =AB + A'CD + B'CD now from here i'm little confused

solution according to calculator is : (A+C)(A+D)(B+C)(B+D)

Answer 1

The proposition initial shows that:

\begin{align} ABCD + AB(CD)' &+ (AB)'CD \\ \left(ABCD + AB(CD)'\right) &+ (AB)'CD \\ AB\left(CD + (CD)'\right) &+ (AB)'CD \\ AB &+ (AB)'CD \\ \left( (AB)'\right. &\left. \ \ \left( \left(AB \right)'CD \right)' \right)' \\ \left( (AB)' \right. & \ \ \left.\left( AB+\left(CD\right)' \right) \right)' \\ \left( (AB)'AB\right. &+\left.(AB)'\left(CD\right)' \right)' \\ \left((AB)'\right. & \ \ \ \ \left.\left(CD\right)' \right)' \\ AB &+ CD \\ \end{align}

But the solution according to the calculator leads to:

\begin{equation} (A+C)(A+D)(B+C)(B+D) \\ (A+AD+AC+CD)(B+BD+BC+CD) \\ \end{equation}

But there is an axiom that $A+AB = A(A+B) = A$, then:

\begin{equation} (A+AD+AC+CD)(B+BD+BC+CD) \\ (A+AC+CD)(B+BC+CD) \\ (A+CD)(B+CD) \\ AB+CD \\ \end{equation}

Then these two answers are the same.

Answer 2

$$ABCD+AB(CD)'+(AB)'CD=AB(CD+(CD)')+(AB)'CD=$$ $$=AB+(AB)'CD=\left((AB)'\left(AB+(CD)'\right)\right)'=\left((AB)'(CD)'\right)'=AB+CD.$$