Question

root 6/root 3 minus root 2

Answer 1

Is it OK!!!
Hope it helps you mate

Answer 2

Heya Here is your answer

$\frac{6}{ \sqrt{3} - \sqrt{2} } \\ rationalizing \: the \: denominator \\ \frac{6}{ \sqrt{3} - \sqrt{2} } \times \frac{ \sqrt{3 } + \sqrt{2} }{ \sqrt{3} + \sqrt{2} } \\ \frac{6( \sqrt{3} - \sqrt{2} ) }{ (\sqrt{3}) {}^{2} - ( \sqrt{2}) {}^{2} } \\ \frac{6( \sqrt{3 } - \sqrt{2} ) }{3 - 2} \\ \frac{6( \sqrt{3} - \sqrt{2} ) }{1 } \\ \frac{6 \sqrt{3} - 6 \sqrt{2} }{1} = 6 \sqrt{3} - 6 \sqrt{2}$
Hope it will help you