Solve 1/3sqrt{27}-sqrt{24}-3sqrt{2/3})*sqrt{12}

$$\frac { 1 } { 3 } \sqrt { 27 } - \sqrt { 24 } - 3 \sqrt { \frac { 2 } { 3 } } ) \cdot \sqrt { 12 }$$

Evaluate

$$\sqrt{3}-3\sqrt{6}\approx -5.616418421$$

Solution Steps

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}.

$$\frac{1}{3}\times 3\sqrt{3}-\sqrt{24}-3\sqrt{\frac{2}{3}}$$

Cancel out 3 and 3.

$$\sqrt{3}-\sqrt{24}-3\sqrt{\frac{2}{3}}$$

Factor 24=2^{2}\times 6. Rewrite the square root of the product \sqrt{2^{2}\times 6} as the product of square roots \sqrt{2^{2}}\sqrt{6}. Take the square root of 2^{2}.

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

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

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

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

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

The square of \sqrt{3} is 3.

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

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

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

Cancel out 3 and 3.

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

Combine -2\sqrt{6} and -\sqrt{6} to get -3\sqrt{6}.

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

Factor

$$\sqrt{3} - 3 \sqrt{6} = -5.616418421$$