Ndtyhij

$A cup B equiv C $ then what is $A$ in terms of $B,C$?

I tried to use $Acup B equiv (A-B)cup(A cap B)cup(B-A) $ to find a similar expression for $A cap B$ but got nowhere.

From the elementary set theory that I did 40 or so years ago I don't recall any material on inverse of set theory operations i.e. union, intersection, complement, difference.

Complement is easy as it is it's own inverse $(A^C)^C=A$

not sure if inverse of difference operator is unique, for example, one can use $(A-B)cup A equiv A$ to construct one inverse of difference $ -X$.

when doing algebra, inverse operations are the first tricks to learn, but was the topic of inverse operations of set theory ever mentioned?

Inverse operators for union, intersection, or set difference are impossible in general.

Knowing $B$ and $C$ is not enough to determine what $A$ is. Consider for example
$$ ? cup {1,2} = {1,2,3} $$
Then either ${1,3}$ or ${2,3}$ would be possible solutions (and there are two more), so there's no operator that given just ${1,2}$ and ${1,2,3}$ can tell you "which of them $A$ really was".

On the other hand symmetric difference
$$ A mathbin{triangle} B = (Asetminus B)cup(Bsetminus A) $$
has an inverse operation, namely itself: $(Amathoptriangle B)mathoptriangle B = A$.

If you consider the algebra of subsets of some universe $U$ under the operations $triangle$ and $cap$, you get a Boolean ring which satisfies the usual ring properties with $triangle$ as addition and $cap$ as multiplication. The ring's $0$ is $varnothing$ and $1$ is $U$ itself.

This gives an opportunity to use more of the usual algebraic rules on sets.
And you can express the remaining set operations in this vocabulary too:
$$ A^complement = 1 mathoptriangle A qquadqquad
Acup B = Amathoptriangle B mathoptriangle(Acap B) $$

What you lose by doing things this way is the nice duality between $cup$ and $cap$ and De Morgan's laws for sets.

(The multiplication still doesn't have an inverse, but it doesn't in general rings either, such as $mathbb Z$).

It's not possible.

Given sets $C$ and $B$ there four states of being we can describe. $x $ in or not in $C$ and $x$ in or not in $B$ and we can only define sets by some combination of those conditions. Those conditions define the following $4$ disjoint basic sets and any set we can possibly be described in terms of $A$ and $B$ will be a union of these sets.

1) ${x not in C, xnot in B} = C^c$ (as $B subset C$)

2) ${x not in C, x in B} = emptyset$ (as $B subset C$)

3) ${x in C, x not in B} = Csetminus B$.

4) ${xin C, x in B} = C cap B = B$ (as $B subset C$).

Although $A$ is disjoint from 1) and 2) and 3) $subset A$ set 4) will typically contain elements in $A$ and as well as elements not in $A$.

Hence in general we can not define $A$ solely on the conditions of whether they are or are not in $C$ or $B$.

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FWIW

We can list all possible sets possible to describe they are

The four above:

1) and [1 and 2] $C^c$, 2)$C^ccap B = emptyset$, 3) and [2 and 3] $Csetminus B$ and 4) and [2 and 4] $B$.

[1 and 3] and [1 and 2 and 3] = $C^c cup Csetminus B = B^c$.

[1 and 4] and [1 and 2 and 4] = $C^c cup B$

[3 and 4] and [2 and 3 and 4] = $(Csetminus B )cup B = C$

[1 and 3 and 4] and [1 and 2 and 3 and 4] = Universal set.