Solve {left(sqrt{6}-sqrt{2}right)}^2+frac{9-sqrt{12}}{sqrt{3}}

$${ \left( \sqrt{ 6 } - \sqrt{ 2 } \right) }^{ 2 } + \frac{ 9- \sqrt{ 12 } }{ \sqrt{ 3 } }$$

Evaluate

$$6-\sqrt{3}\approx 4.267949192$$

Short Solution Steps

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(\sqrt{6}-\sqrt{2}\right)^{2}.

$$\left(\sqrt{6}\right)^{2}-2\sqrt{6}\sqrt{2}+\left(\sqrt{2}\right)^{2}+\frac{9-\sqrt{12}}{\sqrt{3}}$$

The square of \sqrt{6} is 6.

$$6-2\sqrt{6}\sqrt{2}+\left(\sqrt{2}\right)^{2}+\frac{9-\sqrt{12}}{\sqrt{3}}$$

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}.

$$6-2\sqrt{2}\sqrt{3}\sqrt{2}+\left(\sqrt{2}\right)^{2}+\frac{9-\sqrt{12}}{\sqrt{3}}$$

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

$$6-2\times 2\sqrt{3}+\left(\sqrt{2}\right)^{2}+\frac{9-\sqrt{12}}{\sqrt{3}}$$

Multiply -2 and 2 to get -4.

$$6-4\sqrt{3}+\left(\sqrt{2}\right)^{2}+\frac{9-\sqrt{12}}{\sqrt{3}}$$

The square of \sqrt{2} is 2.

$$6-4\sqrt{3}+2+\frac{9-\sqrt{12}}{\sqrt{3}}$$

Add 6 and 2 to get 8.

$$8-4\sqrt{3}+\frac{9-\sqrt{12}}{\sqrt{3}}$$

Factor 12=2^{2}\times 3. Rewrite the square root of the product \sqrt{2^{2}\times 3} as the product of square roots \sqrt{2^{2}}\sqrt{3}. Take the square root of 2^{2}.

$$8-4\sqrt{3}+\frac{9-2\sqrt{3}}{\sqrt{3}}$$

Rationalize the denominator of \frac{9-2\sqrt{3}}{\sqrt{3}} by multiplying numerator and denominator by \sqrt{3}.

$$8-4\sqrt{3}+\frac{\left(9-2\sqrt{3}\right)\sqrt{3}}{\left(\sqrt{3}\right)^{2}}$$

The square of \sqrt{3} is 3.

$$8-4\sqrt{3}+\frac{\left(9-2\sqrt{3}\right)\sqrt{3}}{3}$$

To add or subtract expressions, expand them to make their denominators the same. Multiply 8-4\sqrt{3} times \frac{3}{3}.

$$\frac{3\left(8-4\sqrt{3}\right)}{3}+\frac{\left(9-2\sqrt{3}\right)\sqrt{3}}{3}$$

Since \frac{3\left(8-4\sqrt{3}\right)}{3} and \frac{\left(9-2\sqrt{3}\right)\sqrt{3}}{3} have the same denominator, add them by adding their numerators.

$$\frac{3\left(8-4\sqrt{3}\right)+\left(9-2\sqrt{3}\right)\sqrt{3}}{3}$$

Do the multiplications in 3\left(8-4\sqrt{3}\right)+\left(9-2\sqrt{3}\right)\sqrt{3}.

$$\frac{24-12\sqrt{3}+9\sqrt{3}-6}{3}$$

Do the calculations in 24-12\sqrt{3}+9\sqrt{3}-6.

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

Divide each term of 18-3\sqrt{3} by 3 to get 6-\sqrt{3}.

$$6-\sqrt{3}$$