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$L$ = {$ab,aabb,aaabbb,aaaabbbb....... $}     $\Rightarrow DCFL$
$R$ = {$\epsilon, a ,b , aa,bb,ab,ba,aaa,bbb ,aabb,aaaabbbb,,,,,,,,$}  $\Rightarrow Regular$
$L\cup R$ = {$\epsilon, a ,b , aa,bb,ab,ba,aaa,bbb ,aabb,aaaabbbb,,,,,,,,$} = $R$  $\Rightarrow Regular$

The thing is $R$ is a set of all strings over {$a,b$}. So $union$ of $R$ and $L$ is $R$ only. So it's regular.
$DCFL \cup Regular$ = $DCFL$ and Regular languages are subset of DCFL .
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