Solve frac{sqrt{6}-sqrt{2}}{2}:frac{3sqrt{2}-sqrt{6}}{2}

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

Evaluate

$$\frac{\sqrt{3}}{3}\approx 0.577350269$$

Solution Steps

Divide \frac{\sqrt{6}-\sqrt{2}}{2} by \frac{3\sqrt{2}-\sqrt{6}}{2} by multiplying \frac{\sqrt{6}-\sqrt{2}}{2} by the reciprocal of \frac{3\sqrt{2}-\sqrt{6}}{2}.

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

Cancel out 2 in both numerator and denominator.

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

Rationalize the denominator of \frac{\sqrt{6}-\sqrt{2}}{-\sqrt{6}+3\sqrt{2}} by multiplying numerator and denominator by -\sqrt{6}-3\sqrt{2}.

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

Consider \left(-\sqrt{6}+3\sqrt{2}\right)\left(-\sqrt{6}-3\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}.

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

Expand \left(-\sqrt{6}\right)^{2}.

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

Calculate -1 to the power of 2 and get 1.

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

The square of \sqrt{6} is 6.

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

Multiply 1 and 6 to get 6.

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

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

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

Calculate 3 to the power of 2 and get 9.

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

The square of \sqrt{2} is 2.

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

Multiply 9 and 2 to get 18.

$$\frac{\left(\sqrt{6}-\sqrt{2}\right)\left(-\sqrt{6}-3\sqrt{2}\right)}{6-18}$$

Subtract 18 from 6 to get -12.

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

Apply the distributive property by multiplying each term of \sqrt{6}-\sqrt{2} by each term of -\sqrt{6}-3\sqrt{2}.

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

The square of \sqrt{6} is 6.

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

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}.

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

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

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

Multiply -3 and 2 to get -6.

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

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}.

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

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

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

Combine -6\sqrt{3} and 2\sqrt{3} to get -4\sqrt{3}.

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

The square of \sqrt{2} is 2.

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

Multiply 3 and 2 to get 6.

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

Add -6 and 6 to get 0.

$$\frac{-4\sqrt{3}}{-12}$$

Divide -4\sqrt{3} by -12 to get \frac{1}{3}\sqrt{3}.

$$\frac{1}{3}\sqrt{3}$$