Solve (sqrt{8/27}-(sqrt{3})*sqrt{6}

$$( \sqrt { \frac { 8 } { 27 } } - ( \sqrt { 3 } ) \cdot \sqrt { 6 }$$

Evaluate

$$\frac{2\sqrt{6}}{9}-3\sqrt{2}\approx -3.698309633$$

Short Solution Steps

Rewrite the square root of the division \sqrt{\frac{8}{27}} as the division of square roots \frac{\sqrt{8}}{\sqrt{27}}.

$$\frac{\sqrt{8}}{\sqrt{27}}-\sqrt{3}\sqrt{6}$$

Factor 8=2^{2}\times 2. Rewrite the square root of the product \sqrt{2^{2}\times 2} as the product of square roots \sqrt{2^{2}}\sqrt{2}. Take the square root of 2^{2}.

$$\frac{2\sqrt{2}}{\sqrt{27}}-\sqrt{3}\sqrt{6}$$

Factor 27=3^{2}\times 3. Rewrite the square root of the product \sqrt{3^{2}\times 3} as the product of square roots \sqrt{3^{2}}\sqrt{3}. Take the square root of 3^{2}.

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

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

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

The square of \sqrt{3} is 3.

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

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

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

Multiply 3 and 3 to get 9.

$$\frac{2\sqrt{6}}{9}-\sqrt{3}\sqrt{6}$$

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}.

$$\frac{2\sqrt{6}}{9}-\sqrt{3}\sqrt{3}\sqrt{2}$$

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

$$\frac{2\sqrt{6}}{9}-3\sqrt{2}$$

To add or subtract expressions, expand them to make their denominators the same. Multiply -3\sqrt{2} times \frac{9}{9}.

$$\frac{2\sqrt{6}}{9}+\frac{9\left(-3\right)\sqrt{2}}{9}$$

Since \frac{2\sqrt{6}}{9} and \frac{9\left(-3\right)\sqrt{2}}{9} have the same denominator, add them by adding their numerators.

$$\frac{2\sqrt{6}+9\left(-3\right)\sqrt{2}}{9}$$

Do the multiplications in 2\sqrt{6}+9\left(-3\right)\sqrt{2}.

$$\frac{2\sqrt{6}-27\sqrt{2}}{9}$$