Question : The simplified value of $\frac{3\sqrt 2 }{\sqrt3 + \sqrt6} - \frac{4 \sqrt 3 }{\sqrt{6}+ \sqrt {2}} + \frac{\sqrt 6}{\sqrt{3}+ \sqrt 2}$ is:
Option 1: $\sqrt 2$
Option 2: $\frac{1}{\sqrt2}$
Option 3: $\sqrt 3- \sqrt 2$
Option 4: $0$
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Correct Answer: $0$
Solution : $\frac{3\sqrt 2 }{\sqrt3 + \sqrt6} - \frac{4 \sqrt 3 }{\sqrt{6}+ \sqrt {2}} + \frac{\sqrt 6}{\sqrt{3}+ \sqrt 2}$ $=\frac{3\sqrt 2(\sqrt3 - \sqrt6) }{(\sqrt3 + \sqrt6)(\sqrt3 - \sqrt6)} - \frac{4 \sqrt 3(\sqrt{6}- \sqrt {2}) }{(\sqrt{6}+ \sqrt {2})(\sqrt{6}- \sqrt {2})} + \frac{\sqrt 6(\sqrt{3}- \sqrt 2)}{(\sqrt{3}+ \sqrt 2)(\sqrt{3}- \sqrt 2)}$ $=-\frac{3\sqrt 2(\sqrt3 - \sqrt6) }{3} - \frac{4 \sqrt 3(\sqrt{6}- \sqrt {2}) }{4} + \frac{\sqrt 6(\sqrt{3}- \sqrt 2)}{1}$ $= -\sqrt 2(\sqrt3 - \sqrt6) -\sqrt 3(\sqrt{6}- \sqrt {2}) +\sqrt 6(\sqrt{3}- \sqrt 2)$ $= -\sqrt6 + \sqrt12 -\sqrt{18}+ \sqrt {6} +\sqrt{18}- \sqrt 12$ $=0$ Hence, the correct answer is $0$.
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