Solve 1/sqrt{6+sqrt{3}}

$$\frac { 1 } { \sqrt { 6 } + \sqrt { 3 } }$$

Evaluate

$$\frac{\sqrt{6}-\sqrt{3}}{3}\approx 0.239146312$$

Solution Steps

Rationalize the denominator of \frac{1}{\sqrt{6}+\sqrt{3}} by multiplying numerator and denominator by \sqrt{6}-\sqrt{3}.

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

Consider \left(\sqrt{6}+\sqrt{3}\right)\left(\sqrt{6}-\sqrt{3}\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

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

Square \sqrt{6}. Square \sqrt{3}.

$$\frac{\sqrt{6}-\sqrt{3}}{6-3}$$

Subtract 3 from 6 to get 3.

$$\frac{\sqrt{6}-\sqrt{3}}{3}$$