The expression  $\sqrt{12+6\sqrt{3}}+\sqrt{12-6\sqrt{3}}$  simplifies to 

Let  ${(a+b\sqrt{3})}^2=12+6\sqrt{3}$
&  ${(a-b\sqrt{3})}^2=12-6\sqrt{3}$
Where  $a,b\in Q\Rightarrow a^2=3b^2=12\&2ab=6$
Solving we get,  $a=\sqrt{3}$ (rejected or 3)
 $\sqrt{12+6\sqrt{3}}+\sqrt{12-6\sqrt{3}}=a+b\sqrt{3}+a-b\sqrt{3}=2a=6$