Solve sqrt{(-6)^2}+(-2sqrt{3})^2-sqrt{3}(2sqrt{48}-sqrt{1/3})

$$\sqrt { ( - 6 ) ^ { 2 } } + ( - 2 \sqrt { 3 } ) ^ { 2 } - \sqrt { 3 } ( 2 \sqrt { 48 } - \sqrt { \frac { 1 } { 3 } } )$$

Evaluate

$$-5$$

Short Solution Steps

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

$$\sqrt{36}+\left(-2\sqrt{3}\right)^{2}-\sqrt{3}\left(2\sqrt{48}-\sqrt{\frac{1}{3}}\right)$$

Calculate the square root of 36 and get 6.

$$6+\left(-2\sqrt{3}\right)^{2}-\sqrt{3}\left(2\sqrt{48}-\sqrt{\frac{1}{3}}\right)$$

Expand \left(-2\sqrt{3}\right)^{2}.

$$6+\left(-2\right)^{2}\left(\sqrt{3}\right)^{2}-\sqrt{3}\left(2\sqrt{48}-\sqrt{\frac{1}{3}}\right)$$

Calculate -2 to the power of 2 and get 4.

$$6+4\left(\sqrt{3}\right)^{2}-\sqrt{3}\left(2\sqrt{48}-\sqrt{\frac{1}{3}}\right)$$

The square of \sqrt{3} is 3.

$$6+4\times 3-\sqrt{3}\left(2\sqrt{48}-\sqrt{\frac{1}{3}}\right)$$

Multiply 4 and 3 to get 12.

$$6+12-\sqrt{3}\left(2\sqrt{48}-\sqrt{\frac{1}{3}}\right)$$

Add 6 and 12 to get 18.

$$18-\sqrt{3}\left(2\sqrt{48}-\sqrt{\frac{1}{3}}\right)$$

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}.

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

Multiply 2 and 4 to get 8.

$$18-\sqrt{3}\left(8\sqrt{3}-\sqrt{\frac{1}{3}}\right)$$

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

$$18-\sqrt{3}\left(8\sqrt{3}-\frac{\sqrt{1}}{\sqrt{3}}\right)$$

Calculate the square root of 1 and get 1.

$$18-\sqrt{3}\left(8\sqrt{3}-\frac{1}{\sqrt{3}}\right)$$

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

$$18-\sqrt{3}\left(8\sqrt{3}-\frac{\sqrt{3}}{\left(\sqrt{3}\right)^{2}}\right)$$

The square of \sqrt{3} is 3.

$$18-\sqrt{3}\left(8\sqrt{3}-\frac{\sqrt{3}}{3}\right)$$

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

$$18-\sqrt{3}\left(\frac{3\times 8\sqrt{3}}{3}-\frac{\sqrt{3}}{3}\right)$$

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

$$18-\sqrt{3}\times \frac{3\times 8\sqrt{3}-\sqrt{3}}{3}$$

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

$$18-\sqrt{3}\times \frac{24\sqrt{3}-\sqrt{3}}{3}$$

Do the calculations in 24\sqrt{3}-\sqrt{3}.

$$18-\sqrt{3}\times \frac{23\sqrt{3}}{3}$$

Express \sqrt{3}\times \frac{23\sqrt{3}}{3} as a single fraction.

$$18-\frac{\sqrt{3}\times 23\sqrt{3}}{3}$$

Multiply \sqrt{3} and \sqrt{3} to get 3.

$$18-\frac{3\times 23}{3}$$

Cancel out 3 and 3.

$$18-23$$

Subtract 23 from 18 to get -5.

$$-5$$

Factor

$$-5$$