Solve (sqrt{6}-sqrt{2})^2-frac{sqrt{6}-sqrt{2}}{sqrt{6}+sqrt{2}}

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

Evaluate

$$6-3\sqrt{3}\approx 0.803847577$$

Solution Steps

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

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

The square of \sqrt{6} is 6.

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

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}.

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

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

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

Multiply -2 and 2 to get -4.

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

The square of \sqrt{2} is 2.

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

Add 6 and 2 to get 8.

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

Rationalize the denominator of \frac{\sqrt{6}-\sqrt{2}}{\sqrt{6}+\sqrt{2}} by multiplying numerator and denominator by \sqrt{6}-\sqrt{2}.

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

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

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

Square \sqrt{6}. Square \sqrt{2}.

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

Subtract 2 from 6 to get 4.

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

Multiply \sqrt{6}-\sqrt{2} and \sqrt{6}-\sqrt{2} to get \left(\sqrt{6}-\sqrt{2}\right)^{2}.

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

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

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

The square of \sqrt{6} is 6.

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

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}.

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

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

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

Multiply -2 and 2 to get -4.

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

The square of \sqrt{2} is 2.

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

Add 6 and 2 to get 8.

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

Divide each term of 8-4\sqrt{3} by 4 to get 2-\sqrt{3}.

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

To find the opposite of 2-\sqrt{3}, find the opposite of each term.

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

Subtract 2 from 8 to get 6.

$$6-4\sqrt{3}+\sqrt{3}$$

Combine -4\sqrt{3} and \sqrt{3} to get -3\sqrt{3}.

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

Factor

$$3 {(2 - \sqrt{3})} = 0.803847577$$