Solve (1)(6sqrt{1/27}-2/3sqrt{18})-(sqrt{4/3}-4sqrt{1/2})

$$( 1 ) ( 6 \sqrt { \frac { 1 } { 27 } } - \frac { 2 } { 3 } \sqrt { 18 } ) - ( \sqrt { \frac { 4 } { 3 } } - 4 \sqrt { \frac { 1 } { 2 } } )$$

Evaluate

$$0$$

Short Solution Steps

Rewrite the square root of the division \sqrt{\frac{1}{27}} as the division of square roots \frac{\sqrt{1}}{\sqrt{27}}.

$$1\left(6\times \frac{\sqrt{1}}{\sqrt{27}}-\frac{2}{3}\sqrt{18}\right)-\left(\sqrt{\frac{4}{3}}-4\sqrt{\frac{1}{2}}\right)$$

Calculate the square root of 1 and get 1.

$$1\left(6\times \frac{1}{\sqrt{27}}-\frac{2}{3}\sqrt{18}\right)-\left(\sqrt{\frac{4}{3}}-4\sqrt{\frac{1}{2}}\right)$$

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}.

$$1\left(6\times \frac{1}{3\sqrt{3}}-\frac{2}{3}\sqrt{18}\right)-\left(\sqrt{\frac{4}{3}}-4\sqrt{\frac{1}{2}}\right)$$

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

$$1\left(6\times \frac{\sqrt{3}}{3\left(\sqrt{3}\right)^{2}}-\frac{2}{3}\sqrt{18}\right)-\left(\sqrt{\frac{4}{3}}-4\sqrt{\frac{1}{2}}\right)$$

The square of \sqrt{3} is 3.

$$1\left(6\times \frac{\sqrt{3}}{3\times 3}-\frac{2}{3}\sqrt{18}\right)-\left(\sqrt{\frac{4}{3}}-4\sqrt{\frac{1}{2}}\right)$$

Multiply 3 and 3 to get 9.

$$1\left(6\times \frac{\sqrt{3}}{9}-\frac{2}{3}\sqrt{18}\right)-\left(\sqrt{\frac{4}{3}}-4\sqrt{\frac{1}{2}}\right)$$

Express 6\times \frac{\sqrt{3}}{9} as a single fraction.

$$1\left(\frac{6\sqrt{3}}{9}-\frac{2}{3}\sqrt{18}\right)-\left(\sqrt{\frac{4}{3}}-4\sqrt{\frac{1}{2}}\right)$$

Factor 18=3^{2}\times 2. Rewrite the square root of the product \sqrt{3^{2}\times 2} as the product of square roots \sqrt{3^{2}}\sqrt{2}. Take the square root of 3^{2}.

$$1\left(\frac{6\sqrt{3}}{9}-\frac{2}{3}\times 3\sqrt{2}\right)-\left(\sqrt{\frac{4}{3}}-4\sqrt{\frac{1}{2}}\right)$$

Cancel out 3 and 3.

$$1\left(\frac{6\sqrt{3}}{9}-2\sqrt{2}\right)-\left(\sqrt{\frac{4}{3}}-4\sqrt{\frac{1}{2}}\right)$$

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

$$1\left(\frac{6\sqrt{3}}{9}+\frac{9\left(-2\right)\sqrt{2}}{9}\right)-\left(\sqrt{\frac{4}{3}}-4\sqrt{\frac{1}{2}}\right)$$

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

$$1\times \frac{6\sqrt{3}+9\left(-2\right)\sqrt{2}}{9}-\left(\sqrt{\frac{4}{3}}-4\sqrt{\frac{1}{2}}\right)$$

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

$$1\times \frac{6\sqrt{3}-18\sqrt{2}}{9}-\left(\sqrt{\frac{4}{3}}-4\sqrt{\frac{1}{2}}\right)$$

Express 1\times \frac{6\sqrt{3}-18\sqrt{2}}{9} as a single fraction.

$$\frac{6\sqrt{3}-18\sqrt{2}}{9}-\left(\sqrt{\frac{4}{3}}-4\sqrt{\frac{1}{2}}\right)$$

Rewrite the square root of the division \sqrt{\frac{4}{3}} as the division of square roots \frac{\sqrt{4}}{\sqrt{3}}.

$$\frac{6\sqrt{3}-18\sqrt{2}}{9}-\left(\frac{\sqrt{4}}{\sqrt{3}}-4\sqrt{\frac{1}{2}}\right)$$

Calculate the square root of 4 and get 2.

$$\frac{6\sqrt{3}-18\sqrt{2}}{9}-\left(\frac{2}{\sqrt{3}}-4\sqrt{\frac{1}{2}}\right)$$

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

$$\frac{6\sqrt{3}-18\sqrt{2}}{9}-\left(\frac{2\sqrt{3}}{\left(\sqrt{3}\right)^{2}}-4\sqrt{\frac{1}{2}}\right)$$

The square of \sqrt{3} is 3.

$$\frac{6\sqrt{3}-18\sqrt{2}}{9}-\left(\frac{2\sqrt{3}}{3}-4\sqrt{\frac{1}{2}}\right)$$

Rewrite the square root of the division \sqrt{\frac{1}{2}} as the division of square roots \frac{\sqrt{1}}{\sqrt{2}}.

$$\frac{6\sqrt{3}-18\sqrt{2}}{9}-\left(\frac{2\sqrt{3}}{3}-4\times \frac{\sqrt{1}}{\sqrt{2}}\right)$$

Calculate the square root of 1 and get 1.

$$\frac{6\sqrt{3}-18\sqrt{2}}{9}-\left(\frac{2\sqrt{3}}{3}-4\times \frac{1}{\sqrt{2}}\right)$$

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

$$\frac{6\sqrt{3}-18\sqrt{2}}{9}-\left(\frac{2\sqrt{3}}{3}-4\times \frac{\sqrt{2}}{\left(\sqrt{2}\right)^{2}}\right)$$

The square of \sqrt{2} is 2.

$$\frac{6\sqrt{3}-18\sqrt{2}}{9}-\left(\frac{2\sqrt{3}}{3}-4\times \frac{\sqrt{2}}{2}\right)$$

Cancel out 2, the greatest common factor in 4 and 2.

$$\frac{6\sqrt{3}-18\sqrt{2}}{9}-\left(\frac{2\sqrt{3}}{3}-2\sqrt{2}\right)$$

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

$$\frac{6\sqrt{3}-18\sqrt{2}}{9}-\left(\frac{2\sqrt{3}}{3}+\frac{3\left(-2\right)\sqrt{2}}{3}\right)$$

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

$$\frac{6\sqrt{3}-18\sqrt{2}}{9}-\frac{2\sqrt{3}+3\left(-2\right)\sqrt{2}}{3}$$

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

$$\frac{6\sqrt{3}-18\sqrt{2}}{9}-\frac{2\sqrt{3}-6\sqrt{2}}{3}$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of 9 and 3 is 9. Multiply \frac{2\sqrt{3}-6\sqrt{2}}{3} times \frac{3}{3}.

$$\frac{6\sqrt{3}-18\sqrt{2}}{9}-\frac{3\left(2\sqrt{3}-6\sqrt{2}\right)}{9}$$

Since \frac{6\sqrt{3}-18\sqrt{2}}{9} and \frac{3\left(2\sqrt{3}-6\sqrt{2}\right)}{9} have the same denominator, subtract them by subtracting their numerators.

$$\frac{6\sqrt{3}-18\sqrt{2}-3\left(2\sqrt{3}-6\sqrt{2}\right)}{9}$$

Do the multiplications in 6\sqrt{3}-18\sqrt{2}-3\left(2\sqrt{3}-6\sqrt{2}\right).

$$\frac{6\sqrt{3}-18\sqrt{2}-6\sqrt{3}+18\sqrt{2}}{9}$$

Do the calculations in 6\sqrt{3}-18\sqrt{2}-6\sqrt{3}+18\sqrt{2}.

$$\frac{0}{9}$$

Zero divided by any non-zero number gives zero.

$$0$$

Factor

$$0$$