Use the laws of algebra of sets to show $(A \cup ( B \cap C')) \cap ( A \cup C ) = A$ 
Use the laws of algebra of sets to show that:
$$(A \cup ( B \cap C')) \cap ( A \cup C ) = A$$

Can someone please tell me how to work out such questions and what are the rules that can be used when using laws to prove such quesion.
 A: First using distributivity
$$(A∪(B∩C'))∩(A∪C) = ((A∪B)∩ (A∪C'))∩(A∪C) $$
then associativity
$$ ((A∪B)∩ (A∪C'))∩(A∪C) = \color{teal}{(A∪B)}∩ \color{blue}{(A∪C') ∩ (A∪C)} $$
Can you take it from here?

To show that $\color{teal}{(A∪B)}∩ \color{blue}{(A∪C') ∩ (A∪C)} =A $ we will first show that
$$ \color{blue}{(A∪C')∩(A∪C)} = \color{blue}{A} \tag{1}$$
Then we'll use the absorption law to finish with
$$ \text{LHS} = \color{teal}{(A∪B)} ∩ \color{blue}{A} = A = \text{RHS}. $$
To prove $(1)$ we use distributivity to get:
$$ \color{red}{[A ∩ (A∪C)]} ∪ [C' ∩ (A∪C)] $$
But by absorption law we have
$$ \color{red}{A} ∪ [C' ∩ (A∪C)] $$
But $C' ∩ (A∪C) = (C' ∩ A) ∪ (C' ∩ C) = (C' ∩ A) ∪ \phi = (C' ∩ A) .$ So by absorption again we have $$ \color{red}{A} ∪ [C' ∩ (A∪C)] = \color{red}{A} ∪ (C' ∩ A) = A. $$
QED.
A: By the laws of algebra, I assume you mean the Boolean Algebra axiom for the Boolean Algebra of Sets. Converting into Boolean notation, you would have that $+$ is $\cup$, $\cdot$ is $\cap$, $0 = \emptyset$, and $1 = X$ where $X$ is the universe of the particular boolean algebra 
I will use this frequently, the property of absorption is
$u \cap (u \cup x) = u \text{ and } u \cup (u \cap x) = u$
for all $u$ and $x$. Jech's Set Theory includes it among his axioms, but derives the identity axiom. With the identity axiom (like the definition from wikipedia), you can derive absorption. Depending on what exactly are your boolean algebra axioms, you may or may not need to prove absorption or identity. 
By Distributivity, 
$$(A \cap (A \cup C)) \cup ((B \cap C') \cap (A \cup C))$$
By absorption, 
$A \cap (A \cup C) = A$
So you have
$A \cup ((B \cap C') \cap (A \cup C))$
By associativity of $\cap$
$A \cup (B \cap (C' \cap (A \cup C)))$
By Distributivity
$A \cup (B \cap ((C' \cap A) \cup (C' \cap C)))$
By complementation, $C' \cap C = \emptyset$ so
$A \cup (B \cap (C' \cap A) \cup \emptyset))$
By the identity axiom, $u \cup \emptyset = u$ for any $u$. So 
$A \cup (B \cap (C' \cap A))$
By Distibutivity
$A \cup ((B \cap C') \cap A)$
By absorption
$A$
The proof is complete. 
