# How do you rationalize the denominator and simplify 1/(sqrta+sqrtb+sqrtc)?

Sep 27, 2015

Multiply numerator and denominator by:

$\left(- \sqrt{a} + \sqrt{b} + \sqrt{c}\right) \left(\sqrt{a} - \sqrt{b} + \sqrt{c}\right) \left(\sqrt{a} + \sqrt{b} - \sqrt{c}\right)$

#### Explanation:

For brevity, write:

$\alpha = \sqrt{a}$
$\beta = \sqrt{b}$
$\gamma = \sqrt{c}$

Multiply numerator and denominator by:

$\left(- \alpha + \beta + \gamma\right) \left(\alpha - \beta + \gamma\right) \left(\alpha + \beta - \gamma\right)$

Multiplied out in full, that is:

$- {\alpha}^{3} - {\beta}^{3} - {\gamma}^{3} + {\alpha}^{2} \beta + {\beta}^{2} \gamma + \alpha {\gamma}^{2} + \alpha {\beta}^{2} + \beta {\gamma}^{2} + {\alpha}^{2} \gamma - 2 \alpha \beta \gamma$

The denominator is:

$\left(\alpha + \beta + \gamma\right) \left(- {\alpha}^{3} - {\beta}^{3} - {\gamma}^{3} + {\alpha}^{2} \beta + {\beta}^{2} \gamma + \alpha {\gamma}^{2} + \alpha {\beta}^{2} + \beta {\gamma}^{2} + {\alpha}^{2} \gamma - 2 \alpha \beta \gamma\right)$

$= - {\alpha}^{4} - {\beta}^{4} - {\gamma}^{4} + 2 {\alpha}^{2} {\beta}^{2} + 2 {\beta}^{2} {\gamma}^{2} + 2 {\alpha}^{2} {\gamma}^{2}$

$= 2 a b + 2 b c + 2 a c - {a}^{2} - {b}^{2} - {c}^{2}$

So we can write:

$\frac{1}{\sqrt{a} + \sqrt{b} + \sqrt{c}}$

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

or if you prefer:

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