Question

      6               √6            4√3
________ + ________ - _______
2√3 - √6      √3 + √2      √6 - √2
Simplify

URGENT!!!

Answer 1

$\frac{6}{2 \sqrt{3} - \sqrt{6} } = \frac{6(2 \sqrt{3} + \sqrt{6} )}{(2 \sqrt{3} - \sqrt{6} )(2 \sqrt{3} + \sqrt{6} )} = \frac{6(2 \sqrt{3} + \sqrt{6} )}{12 - 6} = \frac{6(2 \sqrt{3} + \sqrt{6} )}{6} = 2 \sqrt{3} + \sqrt{6}$

$\frac{ \sqrt{6} }{ \sqrt{3} + \sqrt{2} } = \frac{ \sqrt{6}(\sqrt{3} - \sqrt{2}) }{ (\sqrt{3} + \sqrt{2}) (\sqrt{3} - \sqrt{2}) } = \frac{ \sqrt{6}(\sqrt{3} - \sqrt{2}) }{ (3 - 2) } = \frac{ \sqrt{6}(\sqrt{3} - \sqrt{2}) }{ 1 } = \sqrt{6}(\sqrt{3} - \sqrt{2})$

$\frac{4 \sqrt{3} }{ \sqrt{6} - \sqrt{2} } = \frac{4 \sqrt{3} (\sqrt{6} + \sqrt{2})}{ (\sqrt{6} - \sqrt{2})(\sqrt{6} + \sqrt{2}) } = \frac{4 \sqrt{3} (\sqrt{6} + \sqrt{2})}{ (6 - 2) } = \frac{4 \sqrt{3} (\sqrt{6} + \sqrt{2})}{4} = \sqrt{3} (\sqrt{6} + \sqrt{2})$

So,
$(2 \sqrt{3} + \sqrt{6}) + ( \sqrt{6} ( \sqrt{3} - \sqrt{2} )) - ( \sqrt{3} ( \sqrt{6} + \sqrt{2} )) \\ \\ (2 \sqrt{3} + \sqrt{6}) + ( \sqrt{6}\sqrt{3} - \sqrt{6}\sqrt{2} ) - ( \sqrt{3}\sqrt{6} + \sqrt{3}\sqrt{2} ) \\ \\ 2 \sqrt{3} + \sqrt{6} + \sqrt{6}\sqrt{3} - \sqrt{6}\sqrt{2} - \sqrt{3}\sqrt{6} - \sqrt{3}\sqrt{2} \\ \\ 2 \sqrt{3} + \sqrt{6} + \sqrt{6}\sqrt{3} - \sqrt{3}\sqrt{2}\sqrt{2} - \sqrt{3}\sqrt{6} - \sqrt{3}\sqrt{2}$

$2 \sqrt{3} + \sqrt{6} + \sqrt{3}\sqrt{6} - 2\sqrt{3} - \sqrt{3}\sqrt{6} - \sqrt{6}$

You can see each terms is cancelled here.
So, it gives 0 as final sum.


Answer
0