#### Answer

$6-2\sqrt{3}$

#### Work Step by Step

RECALL:
(i) For non-negative real numbers $a$ and $b$,
$\sqrt{a} \cdot \sqrt{b} = \sqrt{ab}$
(ii) For all real numbers $a, b,$ and $c$,
$a(b-c) = ab-ac$
Use rule (ii) above to obtain:
$=\sqrt{6} \cdot \sqrt{6} -\sqrt{6} \cdot \sqrt{2}$
Use rule (i) above to obtain:
$=\sqrt{36} -\sqrt{12}
\\=\sqrt{6^2} -\sqrt{12}
\\=6 -\sqrt{12}$
Factor the radicand so that one of the factors is a perfect square:
$=6-\sqrt{4(3)}
\\=6-\sqrt{2^2(3)}$
Simplify to obtain:
$=6-2\sqrt{3}$