Question

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

Answer 1

Hey friend,
Here is the answer you were looking for:
$\frac{6}{2 \sqrt{3} - \sqrt{6} } + \frac{ \sqrt{6} }{ \sqrt{3} + \sqrt{2} } - \frac{4 \sqrt{3} }{ \sqrt{6} - \sqrt{2} }$

On rationalizing the denominator we get:

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

Using the identity :

$(a + b)(a - b) = {a}^{2} - {b}^{2}$

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

$= \frac{12 \sqrt{3} + 6 \sqrt{6} }{12 - 6} + \frac{ \sqrt{18} - \sqrt{12} }{3 - 2} - \frac{4 \sqrt{18} + 4 \sqrt{6} }{6 - 2} \\ \\ \\ = \frac{12 \sqrt{3} + 6 \sqrt{6} }{6} + \sqrt{2 \times 3 \times 3} - \sqrt{2 \times 2 \times 3} - \frac{4 \sqrt{2 \times 3 \times 3} + 4 \sqrt{6} }{4} \\ \\ \\ = 2 \sqrt{3} + \sqrt{6} + 3 \sqrt{2} - 2 \sqrt{3} - \frac{4 \times 3 \sqrt{2} + 4 \sqrt{6} }{4} \\ \\ \\ = 2 \sqrt{3} + \sqrt{6} + 3 \sqrt{2} - 2 \sqrt{3} - 3 \sqrt{2} - \sqrt{6} \\ \\ = 0$

Hope this helps!!

If you have any doubt regarding to my answer, feel free to ask in the comment section or inbox me if needed.

@Mahak24

Thanks...
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