Let A and B be two sets that $A∩X=B∩X=\phi$ and $A∪X=B∪X$ for some set X. Then,

$A=B$ $A=X$ $B=X$ $A∪B=X$

$A=B$

We have, $A∪X=B∪X$ for some set X $⇒A∩(A∪X)=A∩(B∪X)$ $⇒A=(A∩B)∪(A∩X)$    $[∵A∩(A∪X)=A]$ $⇒A=(A∩B)∪\phi$    $[∵A∩X=\phi]$ $⇒A=A∩B$   ...(i) Again, $A∪X=B∪X$ $⇒B∪(A∪X)=B∩(B∪X)$ $⇒(B∩A)∪(B∩X)=B$ $⇒(B∩A)∪\phi=B$ $⇒A∩B=B$   ...(ii) From (i) and (ii), we have $A = B$