From Förberedande kurs i matematik 1

We can multiply \displaystyle \sqrt{\tfrac{2}{3}} into the bracket and then write the root expressions together under a common root sign using the rule \displaystyle \sqrt{a\vphantom{b}}\cdot \sqrt{b} = \sqrt{ab},

\displaystyle \sqrt{\frac{2}{3}}\bigl(\sqrt{6}-\sqrt{3}\bigr) = \sqrt{\frac{2}{3}}\cdot\sqrt{6} - \sqrt{\frac{2}{3}}\cdot\sqrt{3} = \sqrt{\frac{2\cdot 6}{3}} - \sqrt{\frac{2\cdot 3}{3}}\,\textrm{.}

Because \displaystyle (2\cdot 6)/3 = 2\cdot 2 = 2^2 and \displaystyle (2\cdot 3)/3 = 2, we obtain

\displaystyle \sqrt{\frac{2}{3}}\bigl(\sqrt{6} - \sqrt{3}\bigr) = \sqrt{2^2}-\sqrt{2} = 2-\sqrt{2}\,\textrm{.}