Solve (frac{sqrt{7}+sqrt{6}}{sqrt{3}+2}*frac{sqrt{6}-sqrt{7}}{sqrt{3}-2}):(1/sqrt{3}-frac{sqrt{3}}{9}+1/sqrt{27})

$$( \frac { \sqrt { 7 } + \sqrt { 6 } } { \sqrt { 3 } + 2 } \cdot \frac { \sqrt { 6 } - \sqrt { 7 } } { \sqrt { 3 } - 2 } ) : ( \frac { 1 } { \sqrt { 3 } } - \frac { \sqrt { 3 } } { 9 } + \frac { 1 } { \sqrt { 27 } } )$$

Evaluate

$$\sqrt{3}\approx 1.732050808$$

Solution Steps

Rationalize the denominator of \frac{\sqrt{7}+\sqrt{6}}{\sqrt{3}+2} by multiplying numerator and denominator by \sqrt{3}-2.

$$\frac{\frac{\left(\sqrt{7}+\sqrt{6}\right)\left(\sqrt{3}-2\right)}{\left(\sqrt{3}+2\right)\left(\sqrt{3}-2\right)}\times \frac{\sqrt{6}-\sqrt{7}}{\sqrt{3}-2}}{\frac{1}{\sqrt{3}}-\frac{\sqrt{3}}{9}+\frac{1}{\sqrt{27}}}$$

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

$$\frac{\frac{\left(\sqrt{7}+\sqrt{6}\right)\left(\sqrt{3}-2\right)}{\left(\sqrt{3}\right)^{2}-2^{2}}\times \frac{\sqrt{6}-\sqrt{7}}{\sqrt{3}-2}}{\frac{1}{\sqrt{3}}-\frac{\sqrt{3}}{9}+\frac{1}{\sqrt{27}}}$$

Square \sqrt{3}. Square 2.

$$\frac{\frac{\left(\sqrt{7}+\sqrt{6}\right)\left(\sqrt{3}-2\right)}{3-4}\times \frac{\sqrt{6}-\sqrt{7}}{\sqrt{3}-2}}{\frac{1}{\sqrt{3}}-\frac{\sqrt{3}}{9}+\frac{1}{\sqrt{27}}}$$

Subtract 4 from 3 to get -1.

$$\frac{\frac{\left(\sqrt{7}+\sqrt{6}\right)\left(\sqrt{3}-2\right)}{-1}\times \frac{\sqrt{6}-\sqrt{7}}{\sqrt{3}-2}}{\frac{1}{\sqrt{3}}-\frac{\sqrt{3}}{9}+\frac{1}{\sqrt{27}}}$$

Anything divided by -1 gives its opposite.

$$\frac{\left(-\left(\sqrt{7}+\sqrt{6}\right)\left(\sqrt{3}-2\right)\right)\times \frac{\sqrt{6}-\sqrt{7}}{\sqrt{3}-2}}{\frac{1}{\sqrt{3}}-\frac{\sqrt{3}}{9}+\frac{1}{\sqrt{27}}}$$

Rationalize the denominator of \frac{\sqrt{6}-\sqrt{7}}{\sqrt{3}-2} by multiplying numerator and denominator by \sqrt{3}+2.

$$\frac{\left(-\left(\sqrt{7}+\sqrt{6}\right)\left(\sqrt{3}-2\right)\right)\times \frac{\left(\sqrt{6}-\sqrt{7}\right)\left(\sqrt{3}+2\right)}{\left(\sqrt{3}-2\right)\left(\sqrt{3}+2\right)}}{\frac{1}{\sqrt{3}}-\frac{\sqrt{3}}{9}+\frac{1}{\sqrt{27}}}$$

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

$$\frac{\left(-\left(\sqrt{7}+\sqrt{6}\right)\left(\sqrt{3}-2\right)\right)\times \frac{\left(\sqrt{6}-\sqrt{7}\right)\left(\sqrt{3}+2\right)}{\left(\sqrt{3}\right)^{2}-2^{2}}}{\frac{1}{\sqrt{3}}-\frac{\sqrt{3}}{9}+\frac{1}{\sqrt{27}}}$$

Square \sqrt{3}. Square 2.

$$\frac{\left(-\left(\sqrt{7}+\sqrt{6}\right)\left(\sqrt{3}-2\right)\right)\times \frac{\left(\sqrt{6}-\sqrt{7}\right)\left(\sqrt{3}+2\right)}{3-4}}{\frac{1}{\sqrt{3}}-\frac{\sqrt{3}}{9}+\frac{1}{\sqrt{27}}}$$

Subtract 4 from 3 to get -1.

$$\frac{\left(-\left(\sqrt{7}+\sqrt{6}\right)\left(\sqrt{3}-2\right)\right)\times \frac{\left(\sqrt{6}-\sqrt{7}\right)\left(\sqrt{3}+2\right)}{-1}}{\frac{1}{\sqrt{3}}-\frac{\sqrt{3}}{9}+\frac{1}{\sqrt{27}}}$$

Anything divided by -1 gives its opposite.

$$\frac{\left(-\left(\sqrt{7}+\sqrt{6}\right)\left(\sqrt{3}-2\right)\right)\left(-\left(\sqrt{6}-\sqrt{7}\right)\left(\sqrt{3}+2\right)\right)}{\frac{1}{\sqrt{3}}-\frac{\sqrt{3}}{9}+\frac{1}{\sqrt{27}}}$$

Rationalize the denominator of \frac{1}{\sqrt{3}} by multiplying numerator and denominator by \sqrt{3}.

$$\frac{\left(-\left(\sqrt{7}+\sqrt{6}\right)\left(\sqrt{3}-2\right)\right)\left(-\left(\sqrt{6}-\sqrt{7}\right)\left(\sqrt{3}+2\right)\right)}{\frac{\sqrt{3}}{\left(\sqrt{3}\right)^{2}}-\frac{\sqrt{3}}{9}+\frac{1}{\sqrt{27}}}$$

The square of \sqrt{3} is 3.

$$\frac{\left(-\left(\sqrt{7}+\sqrt{6}\right)\left(\sqrt{3}-2\right)\right)\left(-\left(\sqrt{6}-\sqrt{7}\right)\left(\sqrt{3}+2\right)\right)}{\frac{\sqrt{3}}{3}-\frac{\sqrt{3}}{9}+\frac{1}{\sqrt{27}}}$$

Combine \frac{\sqrt{3}}{3} and -\frac{\sqrt{3}}{9} to get \frac{2}{9}\sqrt{3}.

$$\frac{\left(-\left(\sqrt{7}+\sqrt{6}\right)\left(\sqrt{3}-2\right)\right)\left(-\left(\sqrt{6}-\sqrt{7}\right)\left(\sqrt{3}+2\right)\right)}{\frac{2}{9}\sqrt{3}+\frac{1}{\sqrt{27}}}$$

Factor 27=3^{2}\times 3. Rewrite the square root of the product \sqrt{3^{2}\times 3} as the product of square roots \sqrt{3^{2}}\sqrt{3}. Take the square root of 3^{2}.

$$\frac{\left(-\left(\sqrt{7}+\sqrt{6}\right)\left(\sqrt{3}-2\right)\right)\left(-\left(\sqrt{6}-\sqrt{7}\right)\left(\sqrt{3}+2\right)\right)}{\frac{2}{9}\sqrt{3}+\frac{1}{3\sqrt{3}}}$$

Rationalize the denominator of \frac{1}{3\sqrt{3}} by multiplying numerator and denominator by \sqrt{3}.

$$\frac{\left(-\left(\sqrt{7}+\sqrt{6}\right)\left(\sqrt{3}-2\right)\right)\left(-\left(\sqrt{6}-\sqrt{7}\right)\left(\sqrt{3}+2\right)\right)}{\frac{2}{9}\sqrt{3}+\frac{\sqrt{3}}{3\left(\sqrt{3}\right)^{2}}}$$

The square of \sqrt{3} is 3.

$$\frac{\left(-\left(\sqrt{7}+\sqrt{6}\right)\left(\sqrt{3}-2\right)\right)\left(-\left(\sqrt{6}-\sqrt{7}\right)\left(\sqrt{3}+2\right)\right)}{\frac{2}{9}\sqrt{3}+\frac{\sqrt{3}}{3\times 3}}$$

Multiply 3 and 3 to get 9.

$$\frac{\left(-\left(\sqrt{7}+\sqrt{6}\right)\left(\sqrt{3}-2\right)\right)\left(-\left(\sqrt{6}-\sqrt{7}\right)\left(\sqrt{3}+2\right)\right)}{\frac{2}{9}\sqrt{3}+\frac{\sqrt{3}}{9}}$$

Combine \frac{2}{9}\sqrt{3} and \frac{\sqrt{3}}{9} to get \frac{1}{3}\sqrt{3}.

$$\frac{\left(-\left(\sqrt{7}+\sqrt{6}\right)\left(\sqrt{3}-2\right)\right)\left(-\left(\sqrt{6}-\sqrt{7}\right)\left(\sqrt{3}+2\right)\right)}{\frac{1}{3}\sqrt{3}}$$

Rationalize the denominator of \frac{\left(-\left(\sqrt{7}+\sqrt{6}\right)\left(\sqrt{3}-2\right)\right)\left(-\left(\sqrt{6}-\sqrt{7}\right)\left(\sqrt{3}+2\right)\right)}{\frac{1}{3}\sqrt{3}} by multiplying numerator and denominator by \sqrt{3}.

$$\frac{\left(-\left(\sqrt{7}+\sqrt{6}\right)\left(\sqrt{3}-2\right)\right)\left(-\left(\sqrt{6}-\sqrt{7}\right)\left(\sqrt{3}+2\right)\right)\sqrt{3}}{\frac{1}{3}\left(\sqrt{3}\right)^{2}}$$

The square of \sqrt{3} is 3.

$$\frac{\left(-\left(\sqrt{7}+\sqrt{6}\right)\left(\sqrt{3}-2\right)\right)\left(-\left(\sqrt{6}-\sqrt{7}\right)\left(\sqrt{3}+2\right)\right)\sqrt{3}}{\frac{1}{3}\times 3}$$

Cancel out 3 and 3.

$$\frac{\left(-\left(\sqrt{7}+\sqrt{6}\right)\left(\sqrt{3}-2\right)\right)\left(-\left(\sqrt{6}-\sqrt{7}\right)\left(\sqrt{3}+2\right)\right)\sqrt{3}}{1}$$

Anything divided by one gives itself.

$$\left(-\left(\sqrt{7}+\sqrt{6}\right)\left(\sqrt{3}-2\right)\right)\left(-\left(\sqrt{6}-\sqrt{7}\right)\left(\sqrt{3}+2\right)\right)\sqrt{3}$$

Apply the distributive property by multiplying each term of \sqrt{7}+\sqrt{6} by each term of \sqrt{3}-2.

$$\left(-\left(\sqrt{7}\sqrt{3}-2\sqrt{7}+\sqrt{6}\sqrt{3}-2\sqrt{6}\right)\right)\left(-\left(\sqrt{6}-\sqrt{7}\right)\left(\sqrt{3}+2\right)\right)\sqrt{3}$$

To multiply \sqrt{7} and \sqrt{3}, multiply the numbers under the square root.

$$\left(-\left(\sqrt{21}-2\sqrt{7}+\sqrt{6}\sqrt{3}-2\sqrt{6}\right)\right)\left(-\left(\sqrt{6}-\sqrt{7}\right)\left(\sqrt{3}+2\right)\right)\sqrt{3}$$

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}.

$$\left(-\left(\sqrt{21}-2\sqrt{7}+\sqrt{3}\sqrt{2}\sqrt{3}-2\sqrt{6}\right)\right)\left(-\left(\sqrt{6}-\sqrt{7}\right)\left(\sqrt{3}+2\right)\right)\sqrt{3}$$

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

$$\left(-\left(\sqrt{21}-2\sqrt{7}+3\sqrt{2}-2\sqrt{6}\right)\right)\left(-\left(\sqrt{6}-\sqrt{7}\right)\left(\sqrt{3}+2\right)\right)\sqrt{3}$$

To find the opposite of \sqrt{21}-2\sqrt{7}+3\sqrt{2}-2\sqrt{6}, find the opposite of each term.

$$\left(-\sqrt{21}-\left(-2\sqrt{7}\right)-3\sqrt{2}-\left(-2\sqrt{6}\right)\right)\left(-\left(\sqrt{6}-\sqrt{7}\right)\left(\sqrt{3}+2\right)\right)\sqrt{3}$$

The opposite of -2\sqrt{7} is 2\sqrt{7}.

$$\left(-\sqrt{21}+2\sqrt{7}-3\sqrt{2}-\left(-2\sqrt{6}\right)\right)\left(-\left(\sqrt{6}-\sqrt{7}\right)\left(\sqrt{3}+2\right)\right)\sqrt{3}$$

The opposite of -2\sqrt{6} is 2\sqrt{6}.

$$\left(-\sqrt{21}+2\sqrt{7}-3\sqrt{2}+2\sqrt{6}\right)\left(-\left(\sqrt{6}-\sqrt{7}\right)\left(\sqrt{3}+2\right)\right)\sqrt{3}$$

Apply the distributive property by multiplying each term of \sqrt{6}-\sqrt{7} by each term of \sqrt{3}+2.

$$\left(-\sqrt{21}+2\sqrt{7}-3\sqrt{2}+2\sqrt{6}\right)\left(-\left(\sqrt{6}\sqrt{3}+2\sqrt{6}-\sqrt{7}\sqrt{3}-2\sqrt{7}\right)\right)\sqrt{3}$$

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}.

$$\left(-\sqrt{21}+2\sqrt{7}-3\sqrt{2}+2\sqrt{6}\right)\left(-\left(\sqrt{3}\sqrt{2}\sqrt{3}+2\sqrt{6}-\sqrt{7}\sqrt{3}-2\sqrt{7}\right)\right)\sqrt{3}$$

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

$$\left(-\sqrt{21}+2\sqrt{7}-3\sqrt{2}+2\sqrt{6}\right)\left(-\left(3\sqrt{2}+2\sqrt{6}-\sqrt{7}\sqrt{3}-2\sqrt{7}\right)\right)\sqrt{3}$$

To multiply \sqrt{7} and \sqrt{3}, multiply the numbers under the square root.

$$\left(-\sqrt{21}+2\sqrt{7}-3\sqrt{2}+2\sqrt{6}\right)\left(-\left(3\sqrt{2}+2\sqrt{6}-\sqrt{21}-2\sqrt{7}\right)\right)\sqrt{3}$$

To find the opposite of 3\sqrt{2}+2\sqrt{6}-\sqrt{21}-2\sqrt{7}, find the opposite of each term.

$$\left(-\sqrt{21}+2\sqrt{7}-3\sqrt{2}+2\sqrt{6}\right)\left(-3\sqrt{2}-2\sqrt{6}-\left(-\sqrt{21}\right)-\left(-2\sqrt{7}\right)\right)\sqrt{3}$$

The opposite of -\sqrt{21} is \sqrt{21}.

$$\left(-\sqrt{21}+2\sqrt{7}-3\sqrt{2}+2\sqrt{6}\right)\left(-3\sqrt{2}-2\sqrt{6}+\sqrt{21}-\left(-2\sqrt{7}\right)\right)\sqrt{3}$$

The opposite of -2\sqrt{7} is 2\sqrt{7}.

$$\left(-\sqrt{21}+2\sqrt{7}-3\sqrt{2}+2\sqrt{6}\right)\left(-3\sqrt{2}-2\sqrt{6}+\sqrt{21}+2\sqrt{7}\right)\sqrt{3}$$

Apply the distributive property by multiplying each term of -\sqrt{21}+2\sqrt{7}-3\sqrt{2}+2\sqrt{6} by each term of -3\sqrt{2}-2\sqrt{6}+\sqrt{21}+2\sqrt{7}.

$$\left(3\sqrt{21}\sqrt{2}+2\sqrt{21}\sqrt{6}-\left(\sqrt{21}\right)^{2}-2\sqrt{21}\sqrt{7}-6\sqrt{7}\sqrt{2}-4\sqrt{7}\sqrt{6}+2\sqrt{7}\sqrt{21}+4\left(\sqrt{7}\right)^{2}+9\left(\sqrt{2}\right)^{2}+6\sqrt{2}\sqrt{6}-3\sqrt{2}\sqrt{21}-6\sqrt{7}\sqrt{2}-6\sqrt{6}\sqrt{2}-4\left(\sqrt{6}\right)^{2}+2\sqrt{6}\sqrt{21}+4\sqrt{6}\sqrt{7}\right)\sqrt{3}$$

To multiply \sqrt{21} and \sqrt{2}, multiply the numbers under the square root.

$$\left(3\sqrt{42}+2\sqrt{21}\sqrt{6}-\left(\sqrt{21}\right)^{2}-2\sqrt{21}\sqrt{7}-6\sqrt{7}\sqrt{2}-4\sqrt{7}\sqrt{6}+2\sqrt{7}\sqrt{21}+4\left(\sqrt{7}\right)^{2}+9\left(\sqrt{2}\right)^{2}+6\sqrt{2}\sqrt{6}-3\sqrt{2}\sqrt{21}-6\sqrt{7}\sqrt{2}-6\sqrt{6}\sqrt{2}-4\left(\sqrt{6}\right)^{2}+2\sqrt{6}\sqrt{21}+4\sqrt{6}\sqrt{7}\right)\sqrt{3}$$

To multiply \sqrt{21} and \sqrt{6}, multiply the numbers under the square root.

$$\left(3\sqrt{42}+2\sqrt{126}-\left(\sqrt{21}\right)^{2}-2\sqrt{21}\sqrt{7}-6\sqrt{7}\sqrt{2}-4\sqrt{7}\sqrt{6}+2\sqrt{7}\sqrt{21}+4\left(\sqrt{7}\right)^{2}+9\left(\sqrt{2}\right)^{2}+6\sqrt{2}\sqrt{6}-3\sqrt{2}\sqrt{21}-6\sqrt{7}\sqrt{2}-6\sqrt{6}\sqrt{2}-4\left(\sqrt{6}\right)^{2}+2\sqrt{6}\sqrt{21}+4\sqrt{6}\sqrt{7}\right)\sqrt{3}$$

The square of \sqrt{21} is 21.

$$\left(3\sqrt{42}+2\sqrt{126}-21-2\sqrt{21}\sqrt{7}-6\sqrt{7}\sqrt{2}-4\sqrt{7}\sqrt{6}+2\sqrt{7}\sqrt{21}+4\left(\sqrt{7}\right)^{2}+9\left(\sqrt{2}\right)^{2}+6\sqrt{2}\sqrt{6}-3\sqrt{2}\sqrt{21}-6\sqrt{7}\sqrt{2}-6\sqrt{6}\sqrt{2}-4\left(\sqrt{6}\right)^{2}+2\sqrt{6}\sqrt{21}+4\sqrt{6}\sqrt{7}\right)\sqrt{3}$$

Factor 21=7\times 3. Rewrite the square root of the product \sqrt{7\times 3} as the product of square roots \sqrt{7}\sqrt{3}.

$$\left(3\sqrt{42}+2\sqrt{126}-21-2\sqrt{7}\sqrt{3}\sqrt{7}-6\sqrt{7}\sqrt{2}-4\sqrt{7}\sqrt{6}+2\sqrt{7}\sqrt{21}+4\left(\sqrt{7}\right)^{2}+9\left(\sqrt{2}\right)^{2}+6\sqrt{2}\sqrt{6}-3\sqrt{2}\sqrt{21}-6\sqrt{7}\sqrt{2}-6\sqrt{6}\sqrt{2}-4\left(\sqrt{6}\right)^{2}+2\sqrt{6}\sqrt{21}+4\sqrt{6}\sqrt{7}\right)\sqrt{3}$$

Multiply \sqrt{7} and \sqrt{7} to get 7.

$$\left(3\sqrt{42}+2\sqrt{126}-21-2\times 7\sqrt{3}-6\sqrt{7}\sqrt{2}-4\sqrt{7}\sqrt{6}+2\sqrt{7}\sqrt{21}+4\left(\sqrt{7}\right)^{2}+9\left(\sqrt{2}\right)^{2}+6\sqrt{2}\sqrt{6}-3\sqrt{2}\sqrt{21}-6\sqrt{7}\sqrt{2}-6\sqrt{6}\sqrt{2}-4\left(\sqrt{6}\right)^{2}+2\sqrt{6}\sqrt{21}+4\sqrt{6}\sqrt{7}\right)\sqrt{3}$$

Multiply -2 and 7 to get -14.

$$\left(3\sqrt{42}+2\sqrt{126}-21-14\sqrt{3}-6\sqrt{7}\sqrt{2}-4\sqrt{7}\sqrt{6}+2\sqrt{7}\sqrt{21}+4\left(\sqrt{7}\right)^{2}+9\left(\sqrt{2}\right)^{2}+6\sqrt{2}\sqrt{6}-3\sqrt{2}\sqrt{21}-6\sqrt{7}\sqrt{2}-6\sqrt{6}\sqrt{2}-4\left(\sqrt{6}\right)^{2}+2\sqrt{6}\sqrt{21}+4\sqrt{6}\sqrt{7}\right)\sqrt{3}$$

To multiply \sqrt{7} and \sqrt{2}, multiply the numbers under the square root.

$$\left(3\sqrt{42}+2\sqrt{126}-21-14\sqrt{3}-6\sqrt{14}-4\sqrt{7}\sqrt{6}+2\sqrt{7}\sqrt{21}+4\left(\sqrt{7}\right)^{2}+9\left(\sqrt{2}\right)^{2}+6\sqrt{2}\sqrt{6}-3\sqrt{2}\sqrt{21}-6\sqrt{7}\sqrt{2}-6\sqrt{6}\sqrt{2}-4\left(\sqrt{6}\right)^{2}+2\sqrt{6}\sqrt{21}+4\sqrt{6}\sqrt{7}\right)\sqrt{3}$$

To multiply \sqrt{7} and \sqrt{6}, multiply the numbers under the square root.

$$\left(3\sqrt{42}+2\sqrt{126}-21-14\sqrt{3}-6\sqrt{14}-4\sqrt{42}+2\sqrt{7}\sqrt{21}+4\left(\sqrt{7}\right)^{2}+9\left(\sqrt{2}\right)^{2}+6\sqrt{2}\sqrt{6}-3\sqrt{2}\sqrt{21}-6\sqrt{7}\sqrt{2}-6\sqrt{6}\sqrt{2}-4\left(\sqrt{6}\right)^{2}+2\sqrt{6}\sqrt{21}+4\sqrt{6}\sqrt{7}\right)\sqrt{3}$$

Combine 3\sqrt{42} and -4\sqrt{42} to get -\sqrt{42}.

$$\left(-\sqrt{42}+2\sqrt{126}-21-14\sqrt{3}-6\sqrt{14}+2\sqrt{7}\sqrt{21}+4\left(\sqrt{7}\right)^{2}+9\left(\sqrt{2}\right)^{2}+6\sqrt{2}\sqrt{6}-3\sqrt{2}\sqrt{21}-6\sqrt{7}\sqrt{2}-6\sqrt{6}\sqrt{2}-4\left(\sqrt{6}\right)^{2}+2\sqrt{6}\sqrt{21}+4\sqrt{6}\sqrt{7}\right)\sqrt{3}$$

Factor 21=7\times 3. Rewrite the square root of the product \sqrt{7\times 3} as the product of square roots \sqrt{7}\sqrt{3}.

$$\left(-\sqrt{42}+2\sqrt{126}-21-14\sqrt{3}-6\sqrt{14}+2\sqrt{7}\sqrt{7}\sqrt{3}+4\left(\sqrt{7}\right)^{2}+9\left(\sqrt{2}\right)^{2}+6\sqrt{2}\sqrt{6}-3\sqrt{2}\sqrt{21}-6\sqrt{7}\sqrt{2}-6\sqrt{6}\sqrt{2}-4\left(\sqrt{6}\right)^{2}+2\sqrt{6}\sqrt{21}+4\sqrt{6}\sqrt{7}\right)\sqrt{3}$$

Multiply \sqrt{7} and \sqrt{7} to get 7.

$$\left(-\sqrt{42}+2\sqrt{126}-21-14\sqrt{3}-6\sqrt{14}+2\times 7\sqrt{3}+4\left(\sqrt{7}\right)^{2}+9\left(\sqrt{2}\right)^{2}+6\sqrt{2}\sqrt{6}-3\sqrt{2}\sqrt{21}-6\sqrt{7}\sqrt{2}-6\sqrt{6}\sqrt{2}-4\left(\sqrt{6}\right)^{2}+2\sqrt{6}\sqrt{21}+4\sqrt{6}\sqrt{7}\right)\sqrt{3}$$

Multiply 2 and 7 to get 14.

$$\left(-\sqrt{42}+2\sqrt{126}-21-14\sqrt{3}-6\sqrt{14}+14\sqrt{3}+4\left(\sqrt{7}\right)^{2}+9\left(\sqrt{2}\right)^{2}+6\sqrt{2}\sqrt{6}-3\sqrt{2}\sqrt{21}-6\sqrt{7}\sqrt{2}-6\sqrt{6}\sqrt{2}-4\left(\sqrt{6}\right)^{2}+2\sqrt{6}\sqrt{21}+4\sqrt{6}\sqrt{7}\right)\sqrt{3}$$

Combine -14\sqrt{3} and 14\sqrt{3} to get 0.

$$\left(-\sqrt{42}+2\sqrt{126}-21-6\sqrt{14}+4\left(\sqrt{7}\right)^{2}+9\left(\sqrt{2}\right)^{2}+6\sqrt{2}\sqrt{6}-3\sqrt{2}\sqrt{21}-6\sqrt{7}\sqrt{2}-6\sqrt{6}\sqrt{2}-4\left(\sqrt{6}\right)^{2}+2\sqrt{6}\sqrt{21}+4\sqrt{6}\sqrt{7}\right)\sqrt{3}$$

The square of \sqrt{7} is 7.

$$\left(-\sqrt{42}+2\sqrt{126}-21-6\sqrt{14}+4\times 7+9\left(\sqrt{2}\right)^{2}+6\sqrt{2}\sqrt{6}-3\sqrt{2}\sqrt{21}-6\sqrt{7}\sqrt{2}-6\sqrt{6}\sqrt{2}-4\left(\sqrt{6}\right)^{2}+2\sqrt{6}\sqrt{21}+4\sqrt{6}\sqrt{7}\right)\sqrt{3}$$

Multiply 4 and 7 to get 28.

$$\left(-\sqrt{42}+2\sqrt{126}-21-6\sqrt{14}+28+9\left(\sqrt{2}\right)^{2}+6\sqrt{2}\sqrt{6}-3\sqrt{2}\sqrt{21}-6\sqrt{7}\sqrt{2}-6\sqrt{6}\sqrt{2}-4\left(\sqrt{6}\right)^{2}+2\sqrt{6}\sqrt{21}+4\sqrt{6}\sqrt{7}\right)\sqrt{3}$$

Add -21 and 28 to get 7.

$$\left(-\sqrt{42}+2\sqrt{126}+7-6\sqrt{14}+9\left(\sqrt{2}\right)^{2}+6\sqrt{2}\sqrt{6}-3\sqrt{2}\sqrt{21}-6\sqrt{7}\sqrt{2}-6\sqrt{6}\sqrt{2}-4\left(\sqrt{6}\right)^{2}+2\sqrt{6}\sqrt{21}+4\sqrt{6}\sqrt{7}\right)\sqrt{3}$$

The square of \sqrt{2} is 2.

$$\left(-\sqrt{42}+2\sqrt{126}+7-6\sqrt{14}+9\times 2+6\sqrt{2}\sqrt{6}-3\sqrt{2}\sqrt{21}-6\sqrt{7}\sqrt{2}-6\sqrt{6}\sqrt{2}-4\left(\sqrt{6}\right)^{2}+2\sqrt{6}\sqrt{21}+4\sqrt{6}\sqrt{7}\right)\sqrt{3}$$

Multiply 9 and 2 to get 18.

$$\left(-\sqrt{42}+2\sqrt{126}+7-6\sqrt{14}+18+6\sqrt{2}\sqrt{6}-3\sqrt{2}\sqrt{21}-6\sqrt{7}\sqrt{2}-6\sqrt{6}\sqrt{2}-4\left(\sqrt{6}\right)^{2}+2\sqrt{6}\sqrt{21}+4\sqrt{6}\sqrt{7}\right)\sqrt{3}$$

Add 7 and 18 to get 25.

$$\left(-\sqrt{42}+2\sqrt{126}+25-6\sqrt{14}+6\sqrt{2}\sqrt{6}-3\sqrt{2}\sqrt{21}-6\sqrt{7}\sqrt{2}-6\sqrt{6}\sqrt{2}-4\left(\sqrt{6}\right)^{2}+2\sqrt{6}\sqrt{21}+4\sqrt{6}\sqrt{7}\right)\sqrt{3}$$

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}.

$$\left(-\sqrt{42}+2\sqrt{126}+25-6\sqrt{14}+6\sqrt{2}\sqrt{2}\sqrt{3}-3\sqrt{2}\sqrt{21}-6\sqrt{7}\sqrt{2}-6\sqrt{6}\sqrt{2}-4\left(\sqrt{6}\right)^{2}+2\sqrt{6}\sqrt{21}+4\sqrt{6}\sqrt{7}\right)\sqrt{3}$$

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

$$\left(-\sqrt{42}+2\sqrt{126}+25-6\sqrt{14}+6\times 2\sqrt{3}-3\sqrt{2}\sqrt{21}-6\sqrt{7}\sqrt{2}-6\sqrt{6}\sqrt{2}-4\left(\sqrt{6}\right)^{2}+2\sqrt{6}\sqrt{21}+4\sqrt{6}\sqrt{7}\right)\sqrt{3}$$

Multiply 6 and 2 to get 12.

$$\left(-\sqrt{42}+2\sqrt{126}+25-6\sqrt{14}+12\sqrt{3}-3\sqrt{2}\sqrt{21}-6\sqrt{7}\sqrt{2}-6\sqrt{6}\sqrt{2}-4\left(\sqrt{6}\right)^{2}+2\sqrt{6}\sqrt{21}+4\sqrt{6}\sqrt{7}\right)\sqrt{3}$$

To multiply \sqrt{2} and \sqrt{21}, multiply the numbers under the square root.

$$\left(-\sqrt{42}+2\sqrt{126}+25-6\sqrt{14}+12\sqrt{3}-3\sqrt{42}-6\sqrt{7}\sqrt{2}-6\sqrt{6}\sqrt{2}-4\left(\sqrt{6}\right)^{2}+2\sqrt{6}\sqrt{21}+4\sqrt{6}\sqrt{7}\right)\sqrt{3}$$

Combine -\sqrt{42} and -3\sqrt{42} to get -4\sqrt{42}.

$$\left(-4\sqrt{42}+2\sqrt{126}+25-6\sqrt{14}+12\sqrt{3}-6\sqrt{7}\sqrt{2}-6\sqrt{6}\sqrt{2}-4\left(\sqrt{6}\right)^{2}+2\sqrt{6}\sqrt{21}+4\sqrt{6}\sqrt{7}\right)\sqrt{3}$$

To multiply \sqrt{7} and \sqrt{2}, multiply the numbers under the square root.

$$\left(-4\sqrt{42}+2\sqrt{126}+25-6\sqrt{14}+12\sqrt{3}-6\sqrt{14}-6\sqrt{6}\sqrt{2}-4\left(\sqrt{6}\right)^{2}+2\sqrt{6}\sqrt{21}+4\sqrt{6}\sqrt{7}\right)\sqrt{3}$$

Combine -6\sqrt{14} and -6\sqrt{14} to get -12\sqrt{14}.

$$\left(-4\sqrt{42}+2\sqrt{126}+25-12\sqrt{14}+12\sqrt{3}-6\sqrt{6}\sqrt{2}-4\left(\sqrt{6}\right)^{2}+2\sqrt{6}\sqrt{21}+4\sqrt{6}\sqrt{7}\right)\sqrt{3}$$

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}.

$$\left(-4\sqrt{42}+2\sqrt{126}+25-12\sqrt{14}+12\sqrt{3}-6\sqrt{2}\sqrt{3}\sqrt{2}-4\left(\sqrt{6}\right)^{2}+2\sqrt{6}\sqrt{21}+4\sqrt{6}\sqrt{7}\right)\sqrt{3}$$

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

$$\left(-4\sqrt{42}+2\sqrt{126}+25-12\sqrt{14}+12\sqrt{3}-6\times 2\sqrt{3}-4\left(\sqrt{6}\right)^{2}+2\sqrt{6}\sqrt{21}+4\sqrt{6}\sqrt{7}\right)\sqrt{3}$$

Multiply -6 and 2 to get -12.

$$\left(-4\sqrt{42}+2\sqrt{126}+25-12\sqrt{14}+12\sqrt{3}-12\sqrt{3}-4\left(\sqrt{6}\right)^{2}+2\sqrt{6}\sqrt{21}+4\sqrt{6}\sqrt{7}\right)\sqrt{3}$$

Combine 12\sqrt{3} and -12\sqrt{3} to get 0.

$$\left(-4\sqrt{42}+2\sqrt{126}+25-12\sqrt{14}-4\left(\sqrt{6}\right)^{2}+2\sqrt{6}\sqrt{21}+4\sqrt{6}\sqrt{7}\right)\sqrt{3}$$

The square of \sqrt{6} is 6.

$$\left(-4\sqrt{42}+2\sqrt{126}+25-12\sqrt{14}-4\times 6+2\sqrt{6}\sqrt{21}+4\sqrt{6}\sqrt{7}\right)\sqrt{3}$$

Multiply -4 and 6 to get -24.

$$\left(-4\sqrt{42}+2\sqrt{126}+25-12\sqrt{14}-24+2\sqrt{6}\sqrt{21}+4\sqrt{6}\sqrt{7}\right)\sqrt{3}$$

Subtract 24 from 25 to get 1.

$$\left(-4\sqrt{42}+2\sqrt{126}+1-12\sqrt{14}+2\sqrt{6}\sqrt{21}+4\sqrt{6}\sqrt{7}\right)\sqrt{3}$$

To multiply \sqrt{6} and \sqrt{21}, multiply the numbers under the square root.

$$\left(-4\sqrt{42}+2\sqrt{126}+1-12\sqrt{14}+2\sqrt{126}+4\sqrt{6}\sqrt{7}\right)\sqrt{3}$$

Combine 2\sqrt{126} and 2\sqrt{126} to get 4\sqrt{126}.

$$\left(-4\sqrt{42}+4\sqrt{126}+1-12\sqrt{14}+4\sqrt{6}\sqrt{7}\right)\sqrt{3}$$

To multiply \sqrt{6} and \sqrt{7}, multiply the numbers under the square root.

$$\left(-4\sqrt{42}+4\sqrt{126}+1-12\sqrt{14}+4\sqrt{42}\right)\sqrt{3}$$

Combine -4\sqrt{42} and 4\sqrt{42} to get 0.

$$\left(4\sqrt{126}+1-12\sqrt{14}\right)\sqrt{3}$$

Use the distributive property to multiply 4\sqrt{126}+1-12\sqrt{14} by \sqrt{3}.

$$4\sqrt{126}\sqrt{3}+\sqrt{3}-12\sqrt{14}\sqrt{3}$$

Factor 126=3\times 42. Rewrite the square root of the product \sqrt{3\times 42} as the product of square roots \sqrt{3}\sqrt{42}.

$$4\sqrt{3}\sqrt{42}\sqrt{3}+\sqrt{3}-12\sqrt{14}\sqrt{3}$$

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

$$4\times 3\sqrt{42}+\sqrt{3}-12\sqrt{14}\sqrt{3}$$

Multiply 4 and 3 to get 12.

$$12\sqrt{42}+\sqrt{3}-12\sqrt{14}\sqrt{3}$$

To multiply \sqrt{14} and \sqrt{3}, multiply the numbers under the square root.

$$12\sqrt{42}+\sqrt{3}-12\sqrt{42}$$

Combine 12\sqrt{42} and -12\sqrt{42} to get 0.

$$\sqrt{3}$$