We are given two sets \( A \) and \( B \) such that
\[ A \cap X = \varnothing,\quad B \cap X = \varnothing \] and \[ A \cup X = B \cup X, \] for some set \( X \).
We have to show that \[ A = B. \]
Using the given hint, we write
\[ A = A \cap (A \cup X). \]
Now apply the distributive law:
\[ A \cap (A \cup X) = (A \cap A) \cup (A \cap X). \]
Since \( A \cap A = A \) and \( A \cap X = \varnothing \), we get
\[ A = A \cup \varnothing = A. \]
Hence,
\[ A = (A \cap A) \cup (A \cap X). \]
Similarly, for set \( B \),
\[ B = B \cap (B \cup X). \]
Using distributive law,
\[ B \cap (B \cup X) = (B \cap B) \cup (B \cap X). \]
Since \( B \cap B = B \) and \( B \cap X = \varnothing \),
\[ B = B \cup \varnothing = B. \]
Given that \[ A \cup X = B \cup X, \] taking intersection with \( A \) on both sides, we get
\[ A \cap (A \cup X) = A \cap (B \cup X). \]
Using distributive law on the right side,
\[ A \cap (B \cup X) = (A \cap B) \cup (A \cap X). \]
Since \( A \cap X = \varnothing \),
\[ A = A \cap B. \]
Similarly, intersecting \( B \) with both sides of \( A \cup X = B \cup X \), we get
\[ B = A \cap B. \]
Therefore,
\[ A = A \cap B = B. \]
Hence, it is proved that
\[ \boxed{A = B}. \]