Solve (2sqrt{3}+sqrt{6})(2sqrt{3}-sqrt{6})-sqrt{2}*(sqrt{1/8}-sqrt{6})

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

Evaluate

$$2\sqrt{3}+\frac{11}{2}\approx 8.964101615$$

Short Solution Steps

Consider \left(2\sqrt{3}+\sqrt{6}\right)\left(2\sqrt{3}-\sqrt{6}\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

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

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

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

Calculate 2 to the power of 2 and get 4.

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

The square of \sqrt{3} is 3.

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

Multiply 4 and 3 to get 12.

$$12-\left(\sqrt{6}\right)^{2}-\sqrt{2}\left(\sqrt{\frac{1}{8}}-\sqrt{6}\right)$$

The square of \sqrt{6} is 6.

$$12-6-\sqrt{2}\left(\sqrt{\frac{1}{8}}-\sqrt{6}\right)$$

Subtract 6 from 12 to get 6.

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

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

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

Calculate the square root of 1 and get 1.

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

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}.

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

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

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

The square of \sqrt{2} is 2.

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

Multiply 2 and 2 to get 4.

$$6-\sqrt{2}\left(\frac{\sqrt{2}}{4}-\sqrt{6}\right)$$

To add or subtract expressions, expand them to make their denominators the same. Multiply \sqrt{6} times \frac{4}{4}.

$$6-\sqrt{2}\left(\frac{\sqrt{2}}{4}-\frac{4\sqrt{6}}{4}\right)$$

Since \frac{\sqrt{2}}{4} and \frac{4\sqrt{6}}{4} have the same denominator, subtract them by subtracting their numerators.

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

Express \sqrt{2}\times \frac{\sqrt{2}-4\sqrt{6}}{4} as a single fraction.

$$6-\frac{\sqrt{2}\left(\sqrt{2}-4\sqrt{6}\right)}{4}$$

Use the distributive property to multiply \sqrt{2} by \sqrt{2}-4\sqrt{6}.

$$6-\frac{\left(\sqrt{2}\right)^{2}-4\sqrt{2}\sqrt{6}}{4}$$

The square of \sqrt{2} is 2.

$$6-\frac{2-4\sqrt{2}\sqrt{6}}{4}$$

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

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

Multiply \sqrt{2} and \sqrt{2} to get 2.

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

Multiply -4 and 2 to get -8.

$$6-\frac{2-8\sqrt{3}}{4}$$

To add or subtract expressions, expand them to make their denominators the same. Multiply 6 times \frac{4}{4}.

$$\frac{6\times 4}{4}-\frac{2-8\sqrt{3}}{4}$$

Since \frac{6\times 4}{4} and \frac{2-8\sqrt{3}}{4} have the same denominator, subtract them by subtracting their numerators.

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

Do the multiplications in 6\times 4-\left(2-8\sqrt{3}\right).

$$\frac{24-2+8\sqrt{3}}{4}$$

Do the calculations in 24-2+8\sqrt{3}.

$$\frac{22+8\sqrt{3}}{4}$$