Solve (sqrt{24}-sqrt{1/3})-3(sqrt{1/27}+sqrt{6})

$$( \sqrt { 24 } - \sqrt { \frac { 1 } { 3 } } ) - 3 ( \sqrt { \frac { 1 } { 27 } } + \sqrt { 6 } )$$

Evaluate

$$-\frac{2\sqrt{3}}{3}-\sqrt{6}\approx -3.604190281$$

Short Solution Steps

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}.

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

Rewrite the square root of the division \sqrt{\frac{1}{3}} as the division of square roots \frac{\sqrt{1}}{\sqrt{3}}.

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

Calculate the square root of 1 and get 1.

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

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

$$2\sqrt{6}-\frac{\sqrt{3}}{\left(\sqrt{3}\right)^{2}}-3\left(\sqrt{\frac{1}{27}}+\sqrt{6}\right)$$

The square of \sqrt{3} is 3.

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

To add or subtract expressions, expand them to make their denominators the same. Multiply 2\sqrt{6} times \frac{3}{3}.

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

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

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

Do the multiplications in 3\times 2\sqrt{6}-\sqrt{3}.

$$\frac{6\sqrt{6}-\sqrt{3}}{3}-3\left(\sqrt{\frac{1}{27}}+\sqrt{6}\right)$$

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

$$\frac{6\sqrt{6}-\sqrt{3}}{3}-3\left(\frac{\sqrt{1}}{\sqrt{27}}+\sqrt{6}\right)$$

Calculate the square root of 1 and get 1.

$$\frac{6\sqrt{6}-\sqrt{3}}{3}-3\left(\frac{1}{\sqrt{27}}+\sqrt{6}\right)$$

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}.

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

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

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

The square of \sqrt{3} is 3.

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

Multiply 3 and 3 to get 9.

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

To add or subtract expressions, expand them to make their denominators the same. Multiply \sqrt{6} times \frac{9}{9}.

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

Since \frac{\sqrt{3}}{9} and \frac{9\sqrt{6}}{9} have the same denominator, add them by adding their numerators.

$$\frac{6\sqrt{6}-\sqrt{3}}{3}-3\times \frac{\sqrt{3}+9\sqrt{6}}{9}$$

Cancel out 9, the greatest common factor in 3 and 9.

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

Since \frac{6\sqrt{6}-\sqrt{3}}{3} and \frac{\sqrt{3}+9\sqrt{6}}{3} have the same denominator, subtract them by subtracting their numerators.

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

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

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

Do the calculations in 6\sqrt{6}-\sqrt{3}-\sqrt{3}-9\sqrt{6}.

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