Answer
$3\sqrt{2}-3$
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:
$=\sqrt{3} \cdot \sqrt{6} -\sqrt{3} \cdot \sqrt{3}$
Use rule (i) above to obtain:
$=\sqrt{18} -\sqrt{9}
\\=\sqrt{18} -\sqrt{3^2}
\\=\sqrt{18} -3$
Factor the radicand so that one of the factors is a perfect square:
$=\sqrt{9(2)} -3
\\=\sqrt{3^2(2)} - 3$
Simplify to obtain:
$=3\sqrt{2}-3$
