Simplify : $\frac{\sqrt{6}}{\sqrt{2} + \sqrt{3}} + \frac{3\sqrt{2}}{\sqrt{6 + \sqrt{3}}} - \frac{4\sqrt{3}}{\sqrt{6 + \sqrt{2}}}$

Question

My exams are approaching fast and this question was in one of the sample papers .

I have to simplify $$\frac{\sqrt{6}}{\sqrt{2} + \sqrt{3}} + \frac{3\sqrt{2}}{\sqrt{6 + \sqrt{3}}} - \frac{4\sqrt{3}}{\sqrt{6 + \sqrt{2}}}$$

I am a ninth grader and we have been taught how to rationalize the denominator but I have no idea how I could simplify this one where there is a root of a root . Even my math teacher couldn't do it and said that she'll have to check it out .

Answer 1

Things look pretty nice, if the problem is

$$\frac{\sqrt6}{\sqrt3+\sqrt2}+\frac{3\sqrt2}{\sqrt6+\sqrt3}-\frac{4\sqrt3}{\sqrt6+\sqrt2}$$

Multiplying each term with the conjugate of the respective numerator we get,

$$\frac{\sqrt6(\sqrt3-\sqrt2)}{3-2}+\frac{3\sqrt2(\sqrt6+\sqrt3)}{6-3}-\frac{4\sqrt3(\sqrt6-\sqrt2)}{6-2}$$ $$(\text{ as } (\sqrt3-\sqrt2)(\sqrt3+\sqrt2)=(\sqrt3)^2-(\sqrt2)^2=3-2\text{ and so on} )$$

$$=\sqrt6(\sqrt3-\sqrt2)+\sqrt2(\sqrt6+\sqrt3)-\sqrt3(\sqrt6-\sqrt2)$$

$$=\sqrt{18}-\sqrt{12}+\sqrt{12}-\sqrt6-\sqrt{18}+\sqrt6=0$$