Solve (1)(1+sqrt{3})(sqrt{2}-sqrt{6})-(2sqrt{3}-1)^2:(2)(sqrt{48}-4sqrt{1/8})

$$( 1 ) ( 1 + \sqrt { 3 } ) ( \sqrt { 2 } - \sqrt { 6 } ) - ( 2 \sqrt { 3 } - 1 ) ^ { 2 } : ( 2 ) ( \sqrt { 48 } - 4 \sqrt { \frac { 1 } { 8 } } )$$

Evaluate

$$\frac{9\sqrt{2}}{2}+24-2\sqrt{6}-26\sqrt{3}\approx -19.568339452$$

Short Solution Steps

Use the distributive property to multiply 1 by 1+\sqrt{3}.

$$\left(1+\sqrt{3}\right)\left(\sqrt{2}-\sqrt{6}\right)-\frac{\left(2\sqrt{3}-1\right)^{2}}{2}\left(\sqrt{48}-4\sqrt{\frac{1}{8}}\right)$$

Use the distributive property to multiply 1+\sqrt{3} by \sqrt{2}-\sqrt{6}.

$$\sqrt{2}-\sqrt{6}+\sqrt{3}\sqrt{2}-\sqrt{3}\sqrt{6}-\frac{\left(2\sqrt{3}-1\right)^{2}}{2}\left(\sqrt{48}-4\sqrt{\frac{1}{8}}\right)$$

To multiply \sqrt{3} and \sqrt{2}, multiply the numbers under the square root.

$$\sqrt{2}-\sqrt{6}+\sqrt{6}-\sqrt{3}\sqrt{6}-\frac{\left(2\sqrt{3}-1\right)^{2}}{2}\left(\sqrt{48}-4\sqrt{\frac{1}{8}}\right)$$

Combine -\sqrt{6} and \sqrt{6} to get 0.

$$\sqrt{2}-\sqrt{3}\sqrt{6}-\frac{\left(2\sqrt{3}-1\right)^{2}}{2}\left(\sqrt{48}-4\sqrt{\frac{1}{8}}\right)$$

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

$$\sqrt{2}-\sqrt{3}\sqrt{3}\sqrt{2}-\frac{\left(2\sqrt{3}-1\right)^{2}}{2}\left(\sqrt{48}-4\sqrt{\frac{1}{8}}\right)$$

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

$$\sqrt{2}-3\sqrt{2}-\frac{\left(2\sqrt{3}-1\right)^{2}}{2}\left(\sqrt{48}-4\sqrt{\frac{1}{8}}\right)$$

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

$$-2\sqrt{2}-\frac{\left(2\sqrt{3}-1\right)^{2}}{2}\left(\sqrt{48}-4\sqrt{\frac{1}{8}}\right)$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(2\sqrt{3}-1\right)^{2}.

$$-2\sqrt{2}-\frac{4\left(\sqrt{3}\right)^{2}-4\sqrt{3}+1}{2}\left(\sqrt{48}-4\sqrt{\frac{1}{8}}\right)$$

The square of \sqrt{3} is 3.

$$-2\sqrt{2}-\frac{4\times 3-4\sqrt{3}+1}{2}\left(\sqrt{48}-4\sqrt{\frac{1}{8}}\right)$$

Multiply 4 and 3 to get 12.

$$-2\sqrt{2}-\frac{12-4\sqrt{3}+1}{2}\left(\sqrt{48}-4\sqrt{\frac{1}{8}}\right)$$

Add 12 and 1 to get 13.

$$-2\sqrt{2}-\frac{13-4\sqrt{3}}{2}\left(\sqrt{48}-4\sqrt{\frac{1}{8}}\right)$$

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}.

$$-2\sqrt{2}-\frac{13-4\sqrt{3}}{2}\left(4\sqrt{3}-4\sqrt{\frac{1}{8}}\right)$$

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

$$-2\sqrt{2}-\frac{13-4\sqrt{3}}{2}\left(4\sqrt{3}-4\times \frac{\sqrt{1}}{\sqrt{8}}\right)$$

Calculate the square root of 1 and get 1.

$$-2\sqrt{2}-\frac{13-4\sqrt{3}}{2}\left(4\sqrt{3}-4\times \frac{1}{\sqrt{8}}\right)$$

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}.

$$-2\sqrt{2}-\frac{13-4\sqrt{3}}{2}\left(4\sqrt{3}-4\times \frac{1}{2\sqrt{2}}\right)$$

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

$$-2\sqrt{2}-\frac{13-4\sqrt{3}}{2}\left(4\sqrt{3}-4\times \frac{\sqrt{2}}{2\left(\sqrt{2}\right)^{2}}\right)$$

The square of \sqrt{2} is 2.

$$-2\sqrt{2}-\frac{13-4\sqrt{3}}{2}\left(4\sqrt{3}-4\times \frac{\sqrt{2}}{2\times 2}\right)$$

Multiply 2 and 2 to get 4.

$$-2\sqrt{2}-\frac{13-4\sqrt{3}}{2}\left(4\sqrt{3}-4\times \frac{\sqrt{2}}{4}\right)$$

Cancel out 4 and 4.

$$-2\sqrt{2}-\frac{13-4\sqrt{3}}{2}\left(4\sqrt{3}-\sqrt{2}\right)$$

Express \frac{13-4\sqrt{3}}{2}\left(4\sqrt{3}-\sqrt{2}\right) as a single fraction.

$$-2\sqrt{2}-\frac{\left(13-4\sqrt{3}\right)\left(4\sqrt{3}-\sqrt{2}\right)}{2}$$

Use the distributive property to multiply 13-4\sqrt{3} by 4\sqrt{3}-\sqrt{2}.

$$-2\sqrt{2}-\frac{52\sqrt{3}-13\sqrt{2}-16\left(\sqrt{3}\right)^{2}+4\sqrt{3}\sqrt{2}}{2}$$

The square of \sqrt{3} is 3.

$$-2\sqrt{2}-\frac{52\sqrt{3}-13\sqrt{2}-16\times 3+4\sqrt{3}\sqrt{2}}{2}$$

Multiply -16 and 3 to get -48.

$$-2\sqrt{2}-\frac{52\sqrt{3}-13\sqrt{2}-48+4\sqrt{3}\sqrt{2}}{2}$$

To multiply \sqrt{3} and \sqrt{2}, multiply the numbers under the square root.

$$-2\sqrt{2}-\frac{52\sqrt{3}-13\sqrt{2}-48+4\sqrt{6}}{2}$$

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

$$\frac{2\left(-2\right)\sqrt{2}}{2}-\frac{52\sqrt{3}-13\sqrt{2}-48+4\sqrt{6}}{2}$$

Since \frac{2\left(-2\right)\sqrt{2}}{2} and \frac{52\sqrt{3}-13\sqrt{2}-48+4\sqrt{6}}{2} have the same denominator, subtract them by subtracting their numerators.

$$\frac{2\left(-2\right)\sqrt{2}-\left(52\sqrt{3}-13\sqrt{2}-48+4\sqrt{6}\right)}{2}$$

Do the multiplications in 2\left(-2\right)\sqrt{2}-\left(52\sqrt{3}-13\sqrt{2}-48+4\sqrt{6}\right).

$$\frac{-4\sqrt{2}-52\sqrt{3}+13\sqrt{2}+48-4\sqrt{6}}{2}$$

Do the calculations in -4\sqrt{2}-52\sqrt{3}+13\sqrt{2}+48-4\sqrt{6}.

$$\frac{9\sqrt{2}-52\sqrt{3}+48-4\sqrt{6}}{2}$$

Factor

$$\frac{9 \sqrt{2} + 48 - 4 \sqrt{6} - 52 \sqrt{3}}{2} = -19.568339451678234$$