Here’s the step-by-step solution for the given problems in Boolean Algebra:

(a) Prove that $X \cdot (X + Y) = X$

Solution:

1. Distribute $X$ over $(X + Y)$: $X \cdot (X + Y) = (X \cdot X) + (X \cdot Y)$

2. Apply the Idempotent Law ($X \cdot X = X$): $(X \cdot X) + (X \cdot Y) = X + (X \cdot Y)$

3. Apply Absorption Law ($X + X \cdot Y = X$): $X + (X \cdot Y) = X$

Thus, $X \cdot (X + Y) = X$ is proven.

(b) Prove that $X + X' \cdot Y = X + Y$

Solution:

1. Start with the expression: $X + X' \cdot Y$

2. Apply Absorption Law: $X + X' \cdot Y = (X + Y)$

   • Here, $X$ absorbs $X' \cdot Y$, because $X + Y$ already accounts for all cases.

Thus, $X + X' \cdot Y = X + Y$ is proven.

(c) Simplify and implement $Z = AB + BC(B + C)$

Solution:

Simplify the expression step by step:

1. Expand $BC(B + C)$ using distribution: $BC(B + C) = BC \cdot B + BC \cdot C$

2. Apply Idempotent Law ($B \cdot B = B$) and simplify: $BC \cdot B + BC \cdot C = BC + BC$

   • Since $BC + BC = BC$, we get: $BC(B + C) = BC$

3. Substitute back into $Z$: $Z = AB + BC$

Thus, the simplified form of $Z$ is: $Z = AB + BC$

Final Answers:

1. (a): $X \cdot (X + Y) = X$
2. (b): $X + X' \cdot Y = X + Y$
3. (c): Simplified $Z = AB + BC$

Questions for Further Practice:

1. Prove $X + X \cdot Y = X$.
2. Simplify $A \cdot (A + B) \cdot C$.
3. Verify the expression $X + X' \cdot Y \cdot Z = X + Y \cdot Z$.
4. Use a truth table to prove $X \cdot (X + Y) = X$.
5. Simplify $Z = A + A'B + AB$.

Tip: In Boolean algebra, remember key laws like Idempotent, Absorption, and Distribution to simplify expressions effectively.