# How do you rationalize the denominator and simplify 1 / (1+sqrt3-sqrt5)?

May 10, 2016

$\frac{1}{1 + \sqrt{3} - \sqrt{5}} = \frac{7 + 3 \sqrt{3} + \sqrt{5} + 2 \sqrt{15}}{11}$

#### Explanation:

We will use the difference of squares identity twice:

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

Let us first multiply numerator and denominator by:

$1 + \sqrt{3} + \sqrt{5}$

as follows:

$\frac{1}{1 + \sqrt{3} - \sqrt{5}}$

$= \frac{1 + \sqrt{3} + \sqrt{5}}{\left(1 + \sqrt{3} - \sqrt{5}\right) \left(1 + \sqrt{3} + \sqrt{5}\right)}$

$= \frac{1 + \sqrt{3} + \sqrt{5}}{\left(\left(1 + \sqrt{3}\right) - \sqrt{5}\right) \left(\left(1 + \sqrt{3}\right) + \sqrt{5}\right)}$

$= \frac{1 + \sqrt{3} + \sqrt{5}}{{\left(1 + \sqrt{3}\right)}^{2} - {\left(\sqrt{5}\right)}^{2}}$

$= \frac{1 + \sqrt{3} + \sqrt{5}}{\left(1 + 2 \sqrt{3} + 3\right) - 5}$

$= \frac{1 + \sqrt{3} + \sqrt{5}}{2 \sqrt{3} - 1}$

Then multiply numerator and denominator by $\left(2 \sqrt{3} + 1\right)$ as follows:

$= \frac{\left(1 + \sqrt{3} + \sqrt{5}\right) \left(2 \sqrt{3} + 1\right)}{\left(2 \sqrt{3} - 1\right) \left(2 \sqrt{3} + 1\right)}$

$= \frac{\left(1 + \sqrt{3} + \sqrt{5}\right) \left(2 \sqrt{3} + 1\right)}{{\left(2 \sqrt{3}\right)}^{2} - {1}^{2}}$

$= \frac{1}{11} \left(1 + \sqrt{3} + \sqrt{5}\right) \left(2 \sqrt{3} + 1\right)$

$= \frac{1}{11} \left(2 \sqrt{3} \left(1 + \sqrt{3} + \sqrt{5}\right) + \left(1 + \sqrt{3} + \sqrt{5}\right)\right)$

$= \frac{1}{11} \left(\left(2 \sqrt{3} + 6 + 2 \sqrt{15}\right) + \left(1 + \sqrt{3} + \sqrt{5}\right)\right)$

$= \frac{1}{11} \left(7 + 3 \sqrt{3} + \sqrt{5} + 2 \sqrt{15}\right)$