Solve (sqrt{6}+sqrt{3})(sqrt{2}-sqrt{2/3})=

$$( \sqrt{ 6 } + \sqrt{ 3 } )( \sqrt{ 2 } - \sqrt{ \frac{ 2 }{ 3 } } )=$$

Evaluate

$$\sqrt{6}+2\sqrt{3}-\sqrt{2}-2\approx 2.499377796$$

Short Solution Steps

Rewrite the square root of the division \sqrt{\frac{2}{3}} as the division of square roots \frac{\sqrt{2}}{\sqrt{3}}.

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

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

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

The square of \sqrt{3} is 3.

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

To multiply \sqrt{2} and \sqrt{3}, multiply the numbers under the square root.

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

To add or subtract expressions, expand them to make their denominators the same. Multiply \sqrt{2} times \frac{3}{3}.

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

Since \frac{3\sqrt{2}}{3} and \frac{\sqrt{6}}{3} have the same denominator, subtract them by subtracting their numerators.

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

Express \left(\sqrt{6}+\sqrt{3}\right)\times \frac{3\sqrt{2}-\sqrt{6}}{3} as a single fraction.

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

Apply the distributive property by multiplying each term of \sqrt{6}+\sqrt{3} by each term of 3\sqrt{2}-\sqrt{6}.

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

Factor 6=2\times 3. Rewrite the square root of the product \sqrt{2\times 3} as the product of square roots \sqrt{2}\sqrt{3}.

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

Multiply \sqrt{2} and \sqrt{2} to get 2.

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

Multiply 3 and 2 to get 6.

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

The square of \sqrt{6} is 6.

$$\frac{6\sqrt{3}-6+3\sqrt{3}\sqrt{2}-\sqrt{3}\sqrt{6}}{3}$$

To multiply \sqrt{3} and \sqrt{2}, multiply the numbers under the square root.

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

Factor 6=3\times 2. Rewrite the square root of the product \sqrt{3\times 2} as the product of square roots \sqrt{3}\sqrt{2}.

$$\frac{6\sqrt{3}-6+3\sqrt{6}-\sqrt{3}\sqrt{3}\sqrt{2}}{3}$$

Multiply \sqrt{3} and \sqrt{3} to get 3.

$$\frac{6\sqrt{3}-6+3\sqrt{6}-3\sqrt{2}}{3}$$

Divide each term of 6\sqrt{3}-6+3\sqrt{6}-3\sqrt{2} by 3 to get 2\sqrt{3}-2+\sqrt{6}-\sqrt{2}.

$$2\sqrt{3}-2+\sqrt{6}-\sqrt{2}$$