Are these two identities equivalent (b) $A \setminus (B \setminus C)$.
(c) $(A \setminus B) ∪ (A ∩ C)$
so I begin by writing them using set theory notations
(b) $(x ∈ A) ∧ ¬(x ∈ B ∧ x ∉ C)$
(c) $(x ∈ A ∧ x ∉ B) ∧ (x ∈ A ∧ x ∈ C)$
Now assuming that I wrote the logical translations correctly. The textbook says that these two are equivalent. But I cannot bridge the gap in mind as to how exactly they are equivalent. 
Here's my thought process: in (b) it is possible that there were elements of B that got excluded due to the difference symbol relationship between B and C (B \ C). in (c) the difference between A and B (A \ B) , the set B must then include elements that were not in B in (b) due to (x ∈ B ∧ x ∉ C). 
 A: hint
$$x\in F\backslash G\iff x\in F \land x\notin G $$
The negation is
$$x\notin F\backslash G\iff x\notin F \lor x\in G $$
A: I could prove that the two are equivalent, but if I understand you correctly, you want to know what is wrong with your thought process. So:
Your thought process is correct in so far as that in c) you are indeed excluding more of the $B$'s with $A \setminus B$ in comparison to the $B$'s you are removing in b) ... namely you are now also excluding the elements that are in $B$ and in $C$. However, don't forget that in b) you start with just the elements from A, and so those elements that are in $B$ and in $C$ but not in $A$ are thereby not in $A \setminus B$ either. And those elements that are in $B$ and in $C$ and also in $A$ are added back into c) through the $A \cap C$. And so, ultimately there is no difference after all.
A: Just apply set algebra.  Equivalence of Set Difference, de Morgan's Law, and Distribution.
$A\setminus (B\setminus C) ~{\iff A\cap(B\cap C^\mathsf c)^\mathsf c \\ \iff A \cap (B^\mathsf c\cup C) \\ \iff (A\cap B^\mathsf c)\cup (A\cap C) \\ \iff (A\setminus B)\cup (A\cap C) }$
Or if you must:
$A\setminus (B\setminus C) ~{\iff \{x:x\in A\wedge\neg(x\in B\wedge x\notin C) \}\\ \iff \{x:x\in A \wedge (x\notin B\vee x\in C)\} \\ \iff \{x:(x\in A\wedge x\notin B )\vee (x\in A\wedge x\in C)\} \\ \iff (A\setminus B)\cup (A\cap C) }$

Here's my thought process: in (b) it is possible that there were elements of B that got excluded due to the difference symbol relationship between B and C (B \ C). 

Yes.   $A\setminus (B\setminus C)$ contains everything in A except anything in B that is not also in C. 

in (c) the difference between A and B (A \ B) , the set B must then include elements that were not in B in (b) due to (x ∈ B ∧ x ∉ C). 

Indeed.   $(A\setminus B)\cup(A\cap C)$ contains either everything is A excluding anything also in B, or everything in both A and C.
A: b) is equivalent to $$\{x:x\in A\land x\not\in\{y:y\in B \land y\not\in C\}\}$$
c) is equivalent to $$\{x:x\in A\land x\not\in \{y:y\in B\}\}\cup \{x:x\in A\land x\in \{y:y\in C\}\}$$ using logic, the negation of a negation is the original premise. giving $$\{x:x\in A\land x\not\in \{y:y\in B\}\}\cup \{x:x\in A\land x \not\in \{y:y\not\in C\}\}$$ the parts after the $\land$ combine ( and create an intersection) to give:
$$\{x:x\in A\land x\not\in \{y:y\in B\}\cap\{y:y\not\in C\}\}$$ the intersection contains: $$x\not\in \{y:y\in B\}$$ and $$\{y:y\not\in C\}\}$$  which reduced the whole thing to:
$$\{x:x\in A\land x\not\in \{y:y\in B \land y\not\in C\}\}$$
the exact same as the above.
