I have the expression $\lnot A \land B \land C \lor A \land \lnot B \land C \lor A \land B \land \lnot C \lor A \land B \land C$. This simplifies to $A \land B \lor B \land C \lor C \land A$.
This is what I tried:
1. $\lnot A \land B \land C \lor A \land \lnot B \land C \lor A \land B \land \lnot C \lor A \land B \land C$
2. $A \land (\lnot B \land C \lor B \land \lnot C \lor B \land C) \lor \lnot A \land B \land C$
3. $A \land (B \land (C \lor \lnot C) \lor \lnot B \land C) \lor \lnot A \land B \land C$
4. $A \land (B \lor \lnot B \land C) \lor \lnot A \land B \land C$
5. $A \land B \lor A \land \lnot B \land C \lor \lnot A \land B \land C$
6. $C \land (A \land \lnot B \lor \lnot A \land B) \lor A \land B$
And I don't know what to do next. Maybe there is error in my calculation? If someones spots an error and/or knows what should be my next step, please help.
Here is a useful principle:
Adjacency:
$$(P \land Q) \lor (P \land \neg Q) \Leftrightarrow P$$ If you don't have Adjacency, you can derive it from more basic equivalences:
$$(P \land Q) \lor (P \land \neg Q) \Leftrightarrow \text{Distribution}$$
$$P \land (Q \lor \neg Q) \Leftrightarrow \text{Complement}$$
$$P \land \top \Leftrightarrow \text{Identity}$$
$$P$$
You can use Adjacency at various times, e.g. right from the start you can combine $\neg A \land B \land C$ and $A \land B \land C$ to get $B \land C$
Or, from your line two you can combine $B \land \neg C$ and $B \land C$ to just $B$.
Another useful principle is:
Reduction
$$P \lor (\neg P \land Q) \Leftrightarrow P \lor Q$$
Its proof:
$$P \lor (\neg P \land Q) \Leftrightarrow \text{Distribution}$$
$$(P \lor \neg P) \land (P \lor Q) \Leftrightarrow \text{Complement}$$
$$\top \land (P \lor Q) \Leftrightarrow \text{Identity}$$
$$P \lor Q$$
This one you could use on line 4 to reduce $B \lor (\neg B \land C)$ to $B \lor C$
The reduction principle generalizes to:
Generalized Reduction
$$(P \land R) \lor (\neg P \land Q \land R) \Leftrightarrow (P \land R) \lor (Q \land R)$$
Its proof:
$$(P \land R) \lor (\neg P \land Q \land R) \Leftrightarrow \text{Distribution}$$
$$(P \lor (\neg P \land Q)) \land R \Leftrightarrow \text{Reduction}$$
$$(P \lor Q) \land R \Leftrightarrow \text{Distribution}$$
$$(P \land R) \lor (Q \land R)$$
And this one you could use on line $5$ to reduce $A \land \neg B \land C$ to $A \land C$ in the context of $A \land B$, and to also reduce $\neg A \land B \land C$ to $B \land C$, also in the context of $A \land B$, and that immediately gives you your desired answer.
Putting it all together:
$$(\lnot A \land B \land C) \lor (A \land \lnot B \land C) \lor (A \land B \land \lnot C) \lor (A \land B \land C) \Leftrightarrow \text{Adjacency}$$
$$(B \land C) \lor (A \land \lnot B \land C) \lor (A \land B \land \lnot C) \Leftrightarrow \text{Generalized Reduction x 2}$$
$$(B \land C) \lor (A \land C) \lor (A \land B) \Leftrightarrow \text{Commutation}$$
$$(A \land B) \lor (B \land C) \lor (C \land A)$$