Solve 1/3sqrt{3+3sqrt{2}}

$$\frac{ 1 }{ 3 \sqrt{ 3 } +3 \sqrt{ 2 } }$$

Evaluate

$$\frac{\sqrt{3}-\sqrt{2}}{3}\approx 0.105945748$$

Solution Steps

Rationalize the denominator of \frac{1}{3\sqrt{3}+3\sqrt{2}} by multiplying numerator and denominator by 3\sqrt{3}-3\sqrt{2}.

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

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

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

Expand \left(3\sqrt{3}\right)^{2}.

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

Calculate 3 to the power of 2 and get 9.

$$\frac{3\sqrt{3}-3\sqrt{2}}{9\left(\sqrt{3}\right)^{2}-\left(3\sqrt{2}\right)^{2}}$$

The square of \sqrt{3} is 3.

$$\frac{3\sqrt{3}-3\sqrt{2}}{9\times 3-\left(3\sqrt{2}\right)^{2}}$$

Multiply 9 and 3 to get 27.

$$\frac{3\sqrt{3}-3\sqrt{2}}{27-\left(3\sqrt{2}\right)^{2}}$$

Expand \left(3\sqrt{2}\right)^{2}.

$$\frac{3\sqrt{3}-3\sqrt{2}}{27-3^{2}\left(\sqrt{2}\right)^{2}}$$

Calculate 3 to the power of 2 and get 9.

$$\frac{3\sqrt{3}-3\sqrt{2}}{27-9\left(\sqrt{2}\right)^{2}}$$

The square of \sqrt{2} is 2.

$$\frac{3\sqrt{3}-3\sqrt{2}}{27-9\times 2}$$

Multiply 9 and 2 to get 18.

$$\frac{3\sqrt{3}-3\sqrt{2}}{27-18}$$

Subtract 18 from 27 to get 9.

$$\frac{3\sqrt{3}-3\sqrt{2}}{9}$$

Factor

$$\frac{\sqrt{3} - \sqrt{2}}{3} = 0.10594574839859401$$