Solve 1/3(sqrt{108}-sqrt{41/2}-6sqrt{1/3})-2(sqrt{1/8}-frac{1}{3}sqrt{27})

$$\frac { 1 } { 3 } ( \sqrt { 108 } - \sqrt { 4 \frac { 1 } { 2 } } - 6 \sqrt { \frac { 1 } { 3 } } ) - 2 ( \sqrt { \frac { 1 } { 8 } } - \frac { 1 } { 3 } \sqrt { 27 } )$$

Evaluate

$$\frac{10\sqrt{3}}{3}-\sqrt{2}\approx 4.35928913$$

Short Solution Steps

Factor 108=6^{2}\times 3. Rewrite the square root of the product \sqrt{6^{2}\times 3} as the product of square roots \sqrt{6^{2}}\sqrt{3}. Take the square root of 6^{2}.

$$\frac{1}{3}\left(6\sqrt{3}-\sqrt{\frac{4\times 2+1}{2}}-6\sqrt{\frac{1}{3}}\right)-2\left(\sqrt{\frac{1}{8}}-\frac{1}{3}\sqrt{27}\right)$$

Multiply 4 and 2 to get 8.

$$\frac{1}{3}\left(6\sqrt{3}-\sqrt{\frac{8+1}{2}}-6\sqrt{\frac{1}{3}}\right)-2\left(\sqrt{\frac{1}{8}}-\frac{1}{3}\sqrt{27}\right)$$

Add 8 and 1 to get 9.

$$\frac{1}{3}\left(6\sqrt{3}-\sqrt{\frac{9}{2}}-6\sqrt{\frac{1}{3}}\right)-2\left(\sqrt{\frac{1}{8}}-\frac{1}{3}\sqrt{27}\right)$$

Rewrite the square root of the division \sqrt{\frac{9}{2}} as the division of square roots \frac{\sqrt{9}}{\sqrt{2}}.

$$\frac{1}{3}\left(6\sqrt{3}-\frac{\sqrt{9}}{\sqrt{2}}-6\sqrt{\frac{1}{3}}\right)-2\left(\sqrt{\frac{1}{8}}-\frac{1}{3}\sqrt{27}\right)$$

Calculate the square root of 9 and get 3.

$$\frac{1}{3}\left(6\sqrt{3}-\frac{3}{\sqrt{2}}-6\sqrt{\frac{1}{3}}\right)-2\left(\sqrt{\frac{1}{8}}-\frac{1}{3}\sqrt{27}\right)$$

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

$$\frac{1}{3}\left(6\sqrt{3}-\frac{3\sqrt{2}}{\left(\sqrt{2}\right)^{2}}-6\sqrt{\frac{1}{3}}\right)-2\left(\sqrt{\frac{1}{8}}-\frac{1}{3}\sqrt{27}\right)$$

The square of \sqrt{2} is 2.

$$\frac{1}{3}\left(6\sqrt{3}-\frac{3\sqrt{2}}{2}-6\sqrt{\frac{1}{3}}\right)-2\left(\sqrt{\frac{1}{8}}-\frac{1}{3}\sqrt{27}\right)$$

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

$$\frac{1}{3}\left(6\sqrt{3}-\frac{3\sqrt{2}}{2}-6\times \frac{\sqrt{1}}{\sqrt{3}}\right)-2\left(\sqrt{\frac{1}{8}}-\frac{1}{3}\sqrt{27}\right)$$

Calculate the square root of 1 and get 1.

$$\frac{1}{3}\left(6\sqrt{3}-\frac{3\sqrt{2}}{2}-6\times \frac{1}{\sqrt{3}}\right)-2\left(\sqrt{\frac{1}{8}}-\frac{1}{3}\sqrt{27}\right)$$

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

$$\frac{1}{3}\left(6\sqrt{3}-\frac{3\sqrt{2}}{2}-6\times \frac{\sqrt{3}}{\left(\sqrt{3}\right)^{2}}\right)-2\left(\sqrt{\frac{1}{8}}-\frac{1}{3}\sqrt{27}\right)$$

The square of \sqrt{3} is 3.

$$\frac{1}{3}\left(6\sqrt{3}-\frac{3\sqrt{2}}{2}-6\times \frac{\sqrt{3}}{3}\right)-2\left(\sqrt{\frac{1}{8}}-\frac{1}{3}\sqrt{27}\right)$$

Cancel out 3, the greatest common factor in 6 and 3.

$$\frac{1}{3}\left(6\sqrt{3}-\frac{3\sqrt{2}}{2}-2\sqrt{3}\right)-2\left(\sqrt{\frac{1}{8}}-\frac{1}{3}\sqrt{27}\right)$$

Combine 6\sqrt{3} and -2\sqrt{3} to get 4\sqrt{3}.

$$\frac{1}{3}\left(4\sqrt{3}-\frac{3\sqrt{2}}{2}\right)-2\left(\sqrt{\frac{1}{8}}-\frac{1}{3}\sqrt{27}\right)$$

To add or subtract expressions, expand them to make their denominators the same. Multiply 4\sqrt{3} times \frac{2}{2}.

$$\frac{1}{3}\left(\frac{2\times 4\sqrt{3}}{2}-\frac{3\sqrt{2}}{2}\right)-2\left(\sqrt{\frac{1}{8}}-\frac{1}{3}\sqrt{27}\right)$$

Since \frac{2\times 4\sqrt{3}}{2} and \frac{3\sqrt{2}}{2} have the same denominator, subtract them by subtracting their numerators.

$$\frac{1}{3}\times \frac{2\times 4\sqrt{3}-3\sqrt{2}}{2}-2\left(\sqrt{\frac{1}{8}}-\frac{1}{3}\sqrt{27}\right)$$

Do the multiplications in 2\times 4\sqrt{3}-3\sqrt{2}.

$$\frac{1}{3}\times \frac{8\sqrt{3}-3\sqrt{2}}{2}-2\left(\sqrt{\frac{1}{8}}-\frac{1}{3}\sqrt{27}\right)$$

Multiply \frac{1}{3} times \frac{8\sqrt{3}-3\sqrt{2}}{2} by multiplying numerator times numerator and denominator times denominator.

$$\frac{8\sqrt{3}-3\sqrt{2}}{3\times 2}-2\left(\sqrt{\frac{1}{8}}-\frac{1}{3}\sqrt{27}\right)$$

Rewrite the square root of the division \sqrt{\frac{1}{8}} as the division of square roots \frac{\sqrt{1}}{\sqrt{8}}.

$$\frac{8\sqrt{3}-3\sqrt{2}}{3\times 2}-2\left(\frac{\sqrt{1}}{\sqrt{8}}-\frac{1}{3}\sqrt{27}\right)$$

Calculate the square root of 1 and get 1.

$$\frac{8\sqrt{3}-3\sqrt{2}}{3\times 2}-2\left(\frac{1}{\sqrt{8}}-\frac{1}{3}\sqrt{27}\right)$$

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}.

$$\frac{8\sqrt{3}-3\sqrt{2}}{3\times 2}-2\left(\frac{1}{2\sqrt{2}}-\frac{1}{3}\sqrt{27}\right)$$

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

$$\frac{8\sqrt{3}-3\sqrt{2}}{3\times 2}-2\left(\frac{\sqrt{2}}{2\left(\sqrt{2}\right)^{2}}-\frac{1}{3}\sqrt{27}\right)$$

The square of \sqrt{2} is 2.

$$\frac{8\sqrt{3}-3\sqrt{2}}{3\times 2}-2\left(\frac{\sqrt{2}}{2\times 2}-\frac{1}{3}\sqrt{27}\right)$$

Multiply 2 and 2 to get 4.

$$\frac{8\sqrt{3}-3\sqrt{2}}{3\times 2}-2\left(\frac{\sqrt{2}}{4}-\frac{1}{3}\sqrt{27}\right)$$

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}.

$$\frac{8\sqrt{3}-3\sqrt{2}}{3\times 2}-2\left(\frac{\sqrt{2}}{4}-\frac{1}{3}\times 3\sqrt{3}\right)$$

Cancel out 3 and 3.

$$\frac{8\sqrt{3}-3\sqrt{2}}{3\times 2}-2\left(\frac{\sqrt{2}}{4}-\sqrt{3}\right)$$

To add or subtract expressions, expand them to make their denominators the same. Multiply \sqrt{3} times \frac{4}{4}.

$$\frac{8\sqrt{3}-3\sqrt{2}}{3\times 2}-2\left(\frac{\sqrt{2}}{4}-\frac{4\sqrt{3}}{4}\right)$$

Since \frac{\sqrt{2}}{4} and \frac{4\sqrt{3}}{4} have the same denominator, subtract them by subtracting their numerators.

$$\frac{8\sqrt{3}-3\sqrt{2}}{3\times 2}-2\times \frac{\sqrt{2}-4\sqrt{3}}{4}$$

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

$$\frac{8\sqrt{3}-3\sqrt{2}}{3\times 2}-\frac{\sqrt{2}-4\sqrt{3}}{2}$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of 3\times 2 and 2 is 2\times 3. Multiply \frac{\sqrt{2}-4\sqrt{3}}{2} times \frac{3}{3}.

$$\frac{8\sqrt{3}-3\sqrt{2}}{2\times 3}-\frac{3\left(\sqrt{2}-4\sqrt{3}\right)}{2\times 3}$$

Since \frac{8\sqrt{3}-3\sqrt{2}}{2\times 3} and \frac{3\left(\sqrt{2}-4\sqrt{3}\right)}{2\times 3} have the same denominator, subtract them by subtracting their numerators.

$$\frac{8\sqrt{3}-3\sqrt{2}-3\left(\sqrt{2}-4\sqrt{3}\right)}{2\times 3}$$

Do the multiplications in 8\sqrt{3}-3\sqrt{2}-3\left(\sqrt{2}-4\sqrt{3}\right).

$$\frac{8\sqrt{3}-3\sqrt{2}-3\sqrt{2}+12\sqrt{3}}{2\times 3}$$

Do the calculations in 8\sqrt{3}-3\sqrt{2}-3\sqrt{2}+12\sqrt{3}.

$$\frac{20\sqrt{3}-6\sqrt{2}}{2\times 3}$$

Expand 2\times 3.

$$\frac{20\sqrt{3}-6\sqrt{2}}{6}$$

Factor

$$\frac{10 \sqrt{3} - 3 \sqrt{2}}{3} = 4.3592891295231615$$