Question

Answer 1

Q=\(\dfrac{\left(\dfrac{a-b}{\sqrt{a}+\sqrt{b}}\right)^3+2a\sqrt{a}+b\sqrt{b}}{3a^2+3b\sqrt{ab}}+\dfrac{\sqrt{ab-a}}{a\sqrt{a}-b\sqrt{a}}=\dfrac{\left(\sqrt{a}-\sqrt{b}\right)^3+2a\sqrt{a}+b\sqrt{b}}{3a^2+3b\sqrt{ab}}-\dfrac{\sqrt{a}\left(\sqrt{a}-\sqrt{b}\right)}{\sqrt{a}\left(a-b\right)}\)

 

=\(\dfrac{a\sqrt{a}-b\sqrt{b}-3a\sqrt{b}+3b\sqrt{a}+2a\sqrt{a}+b\sqrt{b}}{3a^2+3b\sqrt{ab}}-\dfrac{1}{\sqrt{a}+\sqrt{b}}=\dfrac{3a\sqrt{a}-3a\sqrt{b}+3b\sqrt{a}}{3a^2+3b\sqrt{ab}}-\dfrac{1}{\sqrt{a}+\sqrt{b}}\)

 

=\(\dfrac{3\sqrt{a}\left(a-\sqrt{ab}+b\right)}{3\sqrt{a}\left(\sqrt{a}+\sqrt{b}\right)\left(a-\sqrt{ab}+b\right)}-\dfrac{1}{\sqrt{a}+\sqrt{b}}=\dfrac{1}{\sqrt{a}+\sqrt{b}}-\dfrac{1}{\sqrt{a}+\sqrt{b}}=0\)