= Solution
By definition the <symmetric difference> is $A\oplus B=(A\setminus B)\cup(B\setminus A)$. Put $D=A\oplus B$ and $E=A\cap B$. Distribution of <set intersection> over <set union> gives
$$
D\cap E=(A\cap B^c\cap A\cap B)\cup(B\cap A^c\cap A\cap B)=\varnothing.
$$
For disjoint <sets>, <set difference> removes nothing, so $D\oplus E=(D\setminus E)\cup(E\setminus D)=D\cup E$. Consequently
$$
\begin{aligned}
(A\oplus B)\oplus(A\cap B)
&=(A\cap B^c)\cup(B\cap A^c)\cup(A\cap B)\\
&=A\cup(B\cap A^c)\\
&=(A\cup B)\cap(A\cup A^c)\\
&=\boxed{A\cup B}.
\end{aligned}
$$
Here the second equality combines $A\cap B^c$ with $A\cap B$; the last uses $A\cup A^c=\mathcal U$.
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