Solve -frac{sqrt{6}}{sqrt{3}+sqrt{2}}+frac{3sqrt{2}}{sqrt{6}+sqrt{3}}-frac{4sqrt{2}}{sqrt{6}+sqrt{2}}

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

Evaluate

$$2\sqrt{3}+2-\sqrt{6}-3\sqrt{2}\approx -1.228028815$$

Solution Steps

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

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

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

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

Square \sqrt{3}. Square \sqrt{2}.

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

Subtract 2 from 3 to get 1.

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

Anything divided by one gives itself.

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

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

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

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

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

Square \sqrt{6}. Square \sqrt{3}.

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

Subtract 3 from 6 to get 3.

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

Rationalize the denominator of \frac{4\sqrt{2}}{\sqrt{6}+\sqrt{2}} by multiplying numerator and denominator by \sqrt{6}-\sqrt{2}.

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

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}.

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

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

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

Subtract 2 from 6 to get 4.

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

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

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

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}.

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

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

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

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}.

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

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

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

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

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

The opposite of -2\sqrt{3} is 2\sqrt{3}.

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

Cancel out 3 and 3.

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

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

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

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}.

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

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

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

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

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

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

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

Cancel out 4 and 4.

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

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

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

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}.

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

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

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

The square of \sqrt{2} is 2.

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

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

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

The opposite of -2 is 2.

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

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

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