# How do you divide (4 + sqrt2)/(3sqrt3 - sqrt6)?

Mar 24, 2015

Simplify the denominator by remembering that
$\left(a - b\right) \times \left(a + b\right) = \left({a}^{2} - {b}^{2}\right)$
and therefore we can get rid of the roots in the denominator by multiplying by $3 \sqrt{3} + \sqrt{6}$

Of course if we multiply the denominator by $3 \sqrt{3} + \sqrt{6}$ we will have to multiply the numerator by that amount as well.

$\frac{4 + \sqrt{2}}{3 \sqrt{3} - \sqrt{6}} \cdot \frac{3 \sqrt{3} + \sqrt{6}}{3 \sqrt{3} + \sqrt{6}}$

$= \frac{4 \left(3 \sqrt{3} + \sqrt{6}\right) + \sqrt{2} \left(3 \sqrt{3} + \sqrt{6}\right)}{27 - 6}$

$= \frac{12 \sqrt{3} + 4 \sqrt{6} + 3 \sqrt{6} + \sqrt{12}}{21}$

$= \frac{12 \sqrt{3} + 7 \sqrt{6} + \sqrt{12}}{21}$

or
$\sqrt{3} \frac{12 + 7 \sqrt{2} + \sqrt{4}}{21}$

$= \sqrt{3} \frac{14 + 7 \sqrt{2}}{21}$

$= 7 \sqrt{3} \cdot \frac{2 + \sqrt{2}}{3}$

or, possibly
$= 7 \frac{2 + \sqrt{2}}{\sqrt{3}}$