Question

Can anyone answer this??

Answer 1

here is your answer....

Answer 2

Hey friend,
Here is the answer you were looking for:
$\frac{3 \sqrt{2} }{ \sqrt{3} + \sqrt{6} } - \frac{4 \sqrt{3} }{ \sqrt{6} + \sqrt{2} } + \frac{ \sqrt{6} }{ \sqrt{2} + \sqrt{3} } \\ \\ on \: rationalizing \: the \: denominator \: we \: get \\ \\ \frac{3 \sqrt{2} }{ \sqrt{3} + \sqrt{6} } \times \frac{ \sqrt{3} - \sqrt{6} }{ \sqrt{3} - \sqrt{6} } - \frac{ 4\sqrt{3} }{ \sqrt{6} + \sqrt{2} } \times \frac{ \sqrt{6} - \sqrt{2} }{ \sqrt{6} - \sqrt{2} } + \frac{ \sqrt{6} }{ \sqrt{2} + \sqrt{3} } \times \frac{ \sqrt{2} - \sqrt{3} }{ \sqrt{2} - \sqrt{3} } \\ \\ using \: the \: identity \\ (a + b)(a - b) = {a}^{2} - {b}^{2} \\ \\ = \frac{3 \sqrt{2} \times \sqrt{3} - 3 \sqrt{2} \times \sqrt{6} }{ {( \sqrt{3} )}^{2} - {( \sqrt{6}) }^{2} } - \frac{4 \sqrt{3} \times \sqrt{6} - 4 \sqrt{3} \times \sqrt{2} }{ {( \sqrt{6}) }^{2} - {( \sqrt{2}) }^{2} } + \frac{ \sqrt{6} \times \sqrt{2} - \sqrt{6} \times \sqrt{3} }{ {( \sqrt{2} )}^{2} - {( \sqrt{3} )}^{2} } \\ \\ = \frac{3 \sqrt{6} - 3 \sqrt{2 \times 2 \times 3} }{3 - 6} - \frac{4 \sqrt{3 \times 3 \times 2} - 4 \sqrt{6} }{6 - 2} + \frac{ \sqrt{3 \times 2 \times 2} - \sqrt{2 \times 3 \times 3} }{2 - 3} \\ \\ = \frac{3 \sqrt{6} - 3 \times 2 \sqrt{3} }{ - 3} - \frac{4 \times 3 \sqrt{2} - 4 \sqrt{6} }{4} + \frac{2 \sqrt{3} -3 \sqrt{2} }{ - 1} \\ \\ = \frac{ - 3 \sqrt{6} + 6 \sqrt{3} }{3} - \frac{12 \sqrt{2} - 4 \sqrt{6} }{4} + \frac{ - \sqrt{3} + 3 \sqrt{2} }{1} \\ \\ = - \sqrt{6} + 2 \sqrt{3} - 3 \sqrt{2} + \sqrt{6} - \sqrt{3} + 3 \sqrt{2} \\ \\ = 2 \sqrt{3} - \sqrt{3} \\ \\ = \sqrt{3}$

Hope this helps!!!

@Mahak24

Thanks....
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