Solve sqrt{(-6)^2}+sqrt{18}-(2sqrt{48}-frac{sqrt{6}}{3})divsqrt{3}

$$\sqrt { ( - 6 ) ^ { 2 } } + \sqrt { 18 } - ( 2 \sqrt { 48 } - \frac { \sqrt { 6 } } { 3 } ) \div \sqrt { 3 }$$

Evaluate

$$\frac{10\sqrt{2}}{3}-2\approx 2.714045208$$

Short Solution Steps

Calculate -6 to the power of 2 and get 36.

$$\sqrt{36}+\sqrt{18}-\frac{2\sqrt{48}-\frac{\sqrt{6}}{3}}{\sqrt{3}}$$

Calculate the square root of 36 and get 6.

$$6+\sqrt{18}-\frac{2\sqrt{48}-\frac{\sqrt{6}}{3}}{\sqrt{3}}$$

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}.

$$6+3\sqrt{2}-\frac{2\sqrt{48}-\frac{\sqrt{6}}{3}}{\sqrt{3}}$$

Factor 48=4^{2}\times 3. Rewrite the square root of the product \sqrt{4^{2}\times 3} as the product of square roots \sqrt{4^{2}}\sqrt{3}. Take the square root of 4^{2}.

$$6+3\sqrt{2}-\frac{2\times 4\sqrt{3}-\frac{\sqrt{6}}{3}}{\sqrt{3}}$$

Multiply 2 and 4 to get 8.

$$6+3\sqrt{2}-\frac{8\sqrt{3}-\frac{\sqrt{6}}{3}}{\sqrt{3}}$$

To add or subtract expressions, expand them to make their denominators the same. Multiply 8\sqrt{3} times \frac{3}{3}.

$$6+3\sqrt{2}-\frac{\frac{3\times 8\sqrt{3}}{3}-\frac{\sqrt{6}}{3}}{\sqrt{3}}$$

Since \frac{3\times 8\sqrt{3}}{3} and \frac{\sqrt{6}}{3} have the same denominator, subtract them by subtracting their numerators.

$$6+3\sqrt{2}-\frac{\frac{3\times 8\sqrt{3}-\sqrt{6}}{3}}{\sqrt{3}}$$

Do the multiplications in 3\times 8\sqrt{3}-\sqrt{6}.

$$6+3\sqrt{2}-\frac{\frac{24\sqrt{3}-\sqrt{6}}{3}}{\sqrt{3}}$$

Express \frac{\frac{24\sqrt{3}-\sqrt{6}}{3}}{\sqrt{3}} as a single fraction.

$$6+3\sqrt{2}-\frac{24\sqrt{3}-\sqrt{6}}{3\sqrt{3}}$$

Rationalize the denominator of \frac{24\sqrt{3}-\sqrt{6}}{3\sqrt{3}} by multiplying numerator and denominator by \sqrt{3}.

$$6+3\sqrt{2}-\frac{\left(24\sqrt{3}-\sqrt{6}\right)\sqrt{3}}{3\left(\sqrt{3}\right)^{2}}$$

The square of \sqrt{3} is 3.

$$6+3\sqrt{2}-\frac{\left(24\sqrt{3}-\sqrt{6}\right)\sqrt{3}}{3\times 3}$$

Multiply 3 and 3 to get 9.

$$6+3\sqrt{2}-\frac{\left(24\sqrt{3}-\sqrt{6}\right)\sqrt{3}}{9}$$

To add or subtract expressions, expand them to make their denominators the same. Multiply 6+3\sqrt{2} times \frac{9}{9}.

$$\frac{9\left(6+3\sqrt{2}\right)}{9}-\frac{\left(24\sqrt{3}-\sqrt{6}\right)\sqrt{3}}{9}$$

Since \frac{9\left(6+3\sqrt{2}\right)}{9} and \frac{\left(24\sqrt{3}-\sqrt{6}\right)\sqrt{3}}{9} have the same denominator, subtract them by subtracting their numerators.

$$\frac{9\left(6+3\sqrt{2}\right)-\left(24\sqrt{3}-\sqrt{6}\right)\sqrt{3}}{9}$$

Do the multiplications in 9\left(6+3\sqrt{2}\right)-\left(24\sqrt{3}-\sqrt{6}\right)\sqrt{3}.

$$\frac{54+27\sqrt{2}-72+3\sqrt{2}}{9}$$

Do the calculations in 54+27\sqrt{2}-72+3\sqrt{2}.

$$\frac{-18+30\sqrt{2}}{9}$$

Factor

$$\frac{2 {(5 \sqrt{2} - 3)}}{3} = 2.714045207910317$$