= Solution
The assertion is \b[true]. Write $A=L(\alpha)$ and $B=L(\beta)$. The semantics of <regular expressions> reduce the claim to
$$
(A^*B^*)^*=(A\cup B)^*,
$$
where juxtaposition is <concatenation of formal languages> and a star is the <Kleene star>.
Every element of $A^*B^*$ is a finite concatenation of members of $A\cup B$, so belongs to $(A\cup B)^*$. Concatenating any finite number of such elements still belongs to $(A\cup B)^*$, proving $(A^*B^*)^*\subseteq(A\cup B)^*$.
Conversely, $A\subseteq A^*B^*$ because one may take a single member of $A$ followed by the empty <string> from $B^*$. Likewise $B\subseteq A^*B^*$ by taking the empty <string> from $A^*$. Hence any finite concatenation of members of $A\cup B$ lies in $(A^*B^*)^*$. The empty <string> belongs to both sides. Thus
$$
\boxed{L((\alpha^*\beta^*)^*)=L((\alpha+\beta)^*).}
$$
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