Lời giải:
\(\frac{\frac{(a-b)^3}{(\sqrt{a}-\sqrt{b})^3}-b\sqrt{b}+2a\sqrt{a}}{a\sqrt{a}-b\sqrt{b}}=\frac{\frac{[(\sqrt{a}-\sqrt{b})(\sqrt{a}+\sqrt{b})]^3}{(\sqrt{a}-\sqrt{b})^3}-b\sqrt{b}+2a\sqrt{a}}{(\sqrt{a}-\sqrt{b})(a+\sqrt{ab}+b)}\)
\(=\frac{(\sqrt{a}+\sqrt{b})^3-b\sqrt{b}+2a\sqrt{a}}{(\sqrt{a}-\sqrt{b})(a+\sqrt{ab}+b)}\)
\(=\frac{a\sqrt{a}+3a\sqrt{b}+3b\sqrt{a}+b\sqrt{b}-b\sqrt{b}+2a\sqrt{a}}{(\sqrt{a}-\sqrt{b})(a+\sqrt{ab}+b)}=\frac{3\sqrt{a}(a+\sqrt{ab}+b)}{(\sqrt{a}-\sqrt{b})(a+\sqrt{ab}+b)}=\frac{3\sqrt{a}}{\sqrt{a}-\sqrt{b}}\)
\(\frac{3a+3\sqrt{ab}}{b-a}=\frac{3\sqrt{a}(\sqrt{a}+\sqrt{b})}{(\sqrt{b}-\sqrt{a})(\sqrt{b}+\sqrt{a})}=\frac{3\sqrt{a}}{\sqrt{b}-\sqrt{a}}\)
Do đó:
\(\frac{\frac{(a-b)^3}{(\sqrt{a}-\sqrt{b})^3}-b\sqrt{b}+2a\sqrt{a}}{a\sqrt{a}-b\sqrt{b}}+\frac{3a+3\sqrt{ab}}{b-a}=\frac{3\sqrt{a}}{\sqrt{a}-\sqrt{b}}+\frac{3\sqrt{a}}{\sqrt{b}-\sqrt{a}}=0\)
Ta có đpcm.