Let $A$ and $B$ be two sets such that $A\cap X=B\cap X=\varphi$ and $A\cup X=B\cup X$ for same set $X$. Then,

Given, that, $A\cup X=B\cup X$

$\Rightarrow A\cap (A\cup X)=A\cap (B\cup X)$

$\Rightarrow (A\cap A)\cup (A\cap X)=(A\cap B)\cup (A\cap X)$

[Using distributive law]

$\Rightarrow A\cup \varphi =(A\cap B)\cup \varphi [\because A\cap X=\varphi ]$

$\Rightarrow A=A\cap B$   $...\left(\mathrm{i}\right)$

Again, consider $A\cup X=B\cup X$

$\Rightarrow B\cap (A\cup X)=B\cap (B\cup X)$

$\Rightarrow (B\cap A)\cup (B\cap X)=(B\cap B)\cup (B\cap X)$

$\Rightarrow (B\cap A)\cup \varphi =B\cup \varphi [B\cap X=\varphi ]$

$\Rightarrow A\cap B=B$   $...\left(\mathrm{ii}\right)$

Thus from Eq. $\left(\mathrm{i}\right)$ and $\left(\mathrm{ii}\right)$, we get

$A=B$