Proof of $(A \cap B) \cup (B \cap C) \cup (C \cap A) = (A \cup B) \cap (B \cup C) \cap (C \cup A)$ Prove that 
$(A \cap B) \cup (B \cap C) \cup (C \cap A) = (A \cup B) \cap (B \cup C) \cap (C \cup A)$
I have noted $\mathcal{A} = x \in A$, $\mathcal{B} = x \in B$, $\mathcal{C} = x \in C$
So in logical terms, the expression is:
$$(\mathcal{A} \land \mathcal{B}) \lor (\mathcal{B} \land \mathcal{C}) \lor (\mathcal{C} \land \mathcal{A})$$
By applying the distributive law two times, the closest I could get is:
$$\Big[(\mathcal{A} \land \mathcal{C}) \lor (\mathcal{A} \lor \mathcal{C})\Big] \land \Big[(\mathcal{B} \lor \mathcal{A}) \land (\mathcal{B} \lor \mathcal{C})\Big]$$
Could you give me some indications on how to properly prove this by working with formulas (not Venn Diagrams)?
 A: You seems to be in the right track. Indeed we just need to do some distributivity:

Distribuitivity of $\cup$: $(X \cap Y) \cup Z = (X \cup Z) \cap (Y \cup Z)$
Distribuitivity of $\cap$: $(X \cup Y) \cap Z = (X \cap Z) \cup (Y \cap Z)$

Then we start by the left hand side $(A \cap B) \cup (B \cap C) \cup (C \cap A)$ and move by $\cup$-distribution (from left to right) also distributing the inner $\cap$ eventually left from the $\cup$-distribution:
$(A \cap B) \cup (B \cap C) \cup (C \cap A) = $
$=(A \cup (B \cap C)) \cap (B \cup (B \cap C)) \cup (C \cap A)$
$= ((A \cup B) \cap (A \cup C)) \cap (B \cup (B \cap C)) \cup (C \cap A)$
$= ((A \cup B) \cap (A \cup C)) \cap ((B \cup B) \cap (B \cup C)) \cup (C \cap A)$
$= ((A \cup B) \cap (A \cup C) \cap (B \cup C)) \cup (C \cap A)$
And now we continue the process with one more $\cup$-distribution (and the other eventually inner distributions) until get the final statement we want.
A: for $x \in (A \cap B) \cup (B \cap C) \cup (C \cap A) $ prove that $x \in (A \cup B) \cap (B \cup C) \cap (C \cup A)$
After that for $x \in (A \cup B) \cap (B \cup C) \cap (C \cup A)$
prove that $x \in (A \cap B) \cup (B \cap C) \cup (C \cap A)$.
Another Method
Use the property
$(A \cap B) \cup C=(A\cup C) \cap (A\cup C)$.
& $(A \cup B) \cap C=(A\cap C) \cup (A\cap C)$
Just use first $ (A \cap B) \cup (B \cap C)=(A\cup (B \cap C)) \cap (B \cup (B \cap C)) $
Now simplify more, then take $(A \cap B) \cup (B \cap C) \cup (C \cap A)$.
Next simplify $(A \cup B) \cap (B \cup C) \cap (C \cup A)$. Show bothof them are same.
