If $A∆B=A∆C$, then $B=C$ If $A\Delta B=A\Delta C$, then $B=C$.
I use the formula  $X∆Y=(X\setminus Y)\cup(Y\setminus X)$ in $A∆B=A∆C$. And further use the formula $X\setminus Y=X\cap Y^c$. I am not  able to reach the conclusion. Please help me.

You can do element chasing.
Or you can manipulate symbols and properties and identities.
But you can also divide and conquer.
Divide the space into $8$ disjoint sets determined by the elements that are are not in $A,B,C$.
They are:
1) $A^c \cap B^c \cap C^c$ (the elements that are in none)
2)  $A^c \cap B^c \cap C$
3)  $A^c \cap B \cap C^c$
4)  $A^c \cap B \cap C$
5)  $A \cap B^c \cap C^c$
6) $A \cap B^c \cap C$
7) $A \cap B \cap C^c$
8) $A \cap B \cap C$
$A\Delta B$ consists of $3$ and $4$ and $5$ and $6$  but none of the others $1,2,7,8$.
$A \Delta C$ consists of $2$ and $4$ and $5$ and $7$ but none of the others $1,3,6,8$.
So we get the seemingly paradoxical result that sets $2,3,6,7$ are all both subsets of $A \Delta B = A \Delta C$, and completely disjoint from $A \Delta B = A \Delta C$.  The only way a set can be both a subset and disjoint from another set, is if the set is empty.  So $2,3,6,7$ are all empty.
$B = 3 \cup 4\cup 7 \cup 8 = 4 \cup 8$.
$C = 2 \cup 4 \cup 6 \cup 8 = 4 \cup 8 = B$.
