# How do you simplify sqrt3/(sqrt6 -1) - sqrt3/(sqrt6 + 1)?

Feb 8, 2016

$\frac{2 \sqrt{3}}{5}$

#### Explanation:

You need to find the least common denominator to be able to subtract the two fractions.

In your case, the least common denominator is $\left(\sqrt{6} - 1\right) \left(\sqrt{6} + 1\right)$, so you need to expand the first fraction with $\left(\sqrt{6} + 1\right)$ and the second one with $\left(\sqrt{6} - 1\right)$.

sqrt(3) / (sqrt(6) - 1) - sqrt(3) / (sqrt(6) + 1) = (sqrt(3)color(blue)((sqrt(6) + 1))) / ((sqrt(6) - 1)color(blue)((sqrt(6) + 1))) - (sqrt(3)color(green)((sqrt(6) - 1))) / ((sqrt(6) + 1)color(green)((sqrt(6) - 1)))

$= \frac{\sqrt{3} \left(\sqrt{6} + 1\right) - \sqrt{3} \left(\sqrt{6} - 1\right)}{\left(\sqrt{6} - 1\right) \left(\sqrt{6} + 1\right)}$

... use the formula $\left(a + b\right) \left(a - b\right) = {a}^{2} - {b}^{2}$ to simplify the denominator...

$= \frac{\sqrt{3} \cdot \sqrt{6} + \sqrt{3} - \sqrt{3} \cdot \sqrt{6} + \sqrt{3}}{{\left(\sqrt{6}\right)}^{2} - {1}^{2}}$

$= \frac{\cancel{\sqrt{3} \cdot \sqrt{6}} + \sqrt{3} - \cancel{\sqrt{3} \cdot \sqrt{6}} + \sqrt{3}}{6 - 1}$

$= \frac{2 \sqrt{3}}{5}$

$= \frac{2}{5} \sqrt{3}$