Lời giải:
$a\sqrt{a}-2b\sqrt{b}-3b\sqrt{a}$
$=(a\sqrt{a}+b\sqrt{b})-3(b\sqrt{b}+b\sqrt{a})$
$=(\sqrt{a}+\sqrt{b})(a-\sqrt{ab}+b)-3b(\sqrt{b}+\sqrt{a})$
$=(\sqrt{a}+\sqrt{b})(a-\sqrt{ab}+b-3b)$
$=(\sqrt{a}+\sqrt{b})(a-\sqrt{ab}-2b)$
$=(\sqrt{a}+\sqrt{b})[(a+\sqrt{ab})-2(\sqrt{ab}+b)]$
$=(\sqrt{a}+\sqrt{b})[\sqrt{a}(\sqrt{a}+\sqrt{b})-2\sqrt{b}(\sqrt{a}+\sqrt{b})]$
$=(\sqrt{a}+\sqrt{b})(\sqrt{a}+\sqrt{b})(\sqrt{a}-2\sqrt{b})$
$=(\sqrt{a}+\sqrt{b})^2(\sqrt{a}-2\sqrt{b})$