# How do you rationalize the denominator and simplify 3/(sqrt[6] – sqrt[3])?

Mar 17, 2016

$\sqrt{6} + \sqrt{3}$

#### Explanation:

Known: ${a}^{2} - {b}^{2} = \left(a - b\right) \left(a + b\right)$

Using this principle:

Multiply by 1 but in the form of $1 = \frac{\sqrt{6} + \sqrt{3}}{\sqrt{6} + \sqrt{3}}$

(3(sqrt(6)+sqrt(3)))/((sqrt(6)-sqrt(3))times(sqrt(6)+sqrt(3))

$= \frac{3 \sqrt{6} + 3 \sqrt{3}}{{\left(\sqrt{6}\right)}^{2} - {\left(\sqrt{3}\right)}^{2}}$

$= \frac{\cancel{3} \sqrt{6} + \cancel{3} \sqrt{3}}{\cancel{3}}$

$= \sqrt{6} + \sqrt{3}$