This is an unusual problem; it is the opposite of simplifying.
Usually, we do not want radicals in the denominator . . nor denominators under a radical.
We have: \(\displaystyle \;\dfrac{2\sqrt{6}}{3} \;=\;\dfrac{2\cdot\sqrt{2}\cdot\sqrt{3}}{3}\)
Multiply by \(\displaystyle \frac{\sqrt{3}}{\sqrt{3}}\!:\;\;\dfrac{2\cdot\sqrt{2}\cdot\sqrt{3}}{3}\cdot\dfrac{\sqrt{3}}{\sqrt{3}} \;=\;\dfrac{2\cdot\sqrt{2}\cdot3}{3\sqrt{3}}\). .
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