Solve (frac{3+sqrt{6}}{sqrt{2}+sqrt{3}}+frac{sqrt{15}-sqrt{3}}{sqrt{5}-1}-sqrt{27}+2)^2(2+sqrt{3})^2

$$( \frac { 3 + \sqrt { 6 } } { \sqrt { 2 } + \sqrt { 3 } } + \frac { \sqrt { 15 } - \sqrt { 3 } } { \sqrt { 5 } - 1 } - \sqrt { 27 } + 2 ) ^ { 2 } ( 2 + \sqrt { 3 } ) ^ { 2 }$$

Evaluate

$$1$$

Solution Steps

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

$$\left(\frac{\left(3+\sqrt{6}\right)\left(\sqrt{2}-\sqrt{3}\right)}{\left(\sqrt{2}+\sqrt{3}\right)\left(\sqrt{2}-\sqrt{3}\right)}+\frac{\sqrt{15}-\sqrt{3}}{\sqrt{5}-1}-\sqrt{27}+2\right)^{2}\left(2+\sqrt{3}\right)^{2}$$

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

$$\left(\frac{\left(3+\sqrt{6}\right)\left(\sqrt{2}-\sqrt{3}\right)}{\left(\sqrt{2}\right)^{2}-\left(\sqrt{3}\right)^{2}}+\frac{\sqrt{15}-\sqrt{3}}{\sqrt{5}-1}-\sqrt{27}+2\right)^{2}\left(2+\sqrt{3}\right)^{2}$$

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

$$\left(\frac{\left(3+\sqrt{6}\right)\left(\sqrt{2}-\sqrt{3}\right)}{2-3}+\frac{\sqrt{15}-\sqrt{3}}{\sqrt{5}-1}-\sqrt{27}+2\right)^{2}\left(2+\sqrt{3}\right)^{2}$$

Subtract 3 from 2 to get -1.

$$\left(\frac{\left(3+\sqrt{6}\right)\left(\sqrt{2}-\sqrt{3}\right)}{-1}+\frac{\sqrt{15}-\sqrt{3}}{\sqrt{5}-1}-\sqrt{27}+2\right)^{2}\left(2+\sqrt{3}\right)^{2}$$

Anything divided by -1 gives its opposite.

$$\left(-\left(3+\sqrt{6}\right)\left(\sqrt{2}-\sqrt{3}\right)+\frac{\sqrt{15}-\sqrt{3}}{\sqrt{5}-1}-\sqrt{27}+2\right)^{2}\left(2+\sqrt{3}\right)^{2}$$

Rationalize the denominator of \frac{\sqrt{15}-\sqrt{3}}{\sqrt{5}-1} by multiplying numerator and denominator by \sqrt{5}+1.

$$\left(-\left(3+\sqrt{6}\right)\left(\sqrt{2}-\sqrt{3}\right)+\frac{\left(\sqrt{15}-\sqrt{3}\right)\left(\sqrt{5}+1\right)}{\left(\sqrt{5}-1\right)\left(\sqrt{5}+1\right)}-\sqrt{27}+2\right)^{2}\left(2+\sqrt{3}\right)^{2}$$

Consider \left(\sqrt{5}-1\right)\left(\sqrt{5}+1\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

$$\left(-\left(3+\sqrt{6}\right)\left(\sqrt{2}-\sqrt{3}\right)+\frac{\left(\sqrt{15}-\sqrt{3}\right)\left(\sqrt{5}+1\right)}{\left(\sqrt{5}\right)^{2}-1^{2}}-\sqrt{27}+2\right)^{2}\left(2+\sqrt{3}\right)^{2}$$

Square \sqrt{5}. Square 1.

$$\left(-\left(3+\sqrt{6}\right)\left(\sqrt{2}-\sqrt{3}\right)+\frac{\left(\sqrt{15}-\sqrt{3}\right)\left(\sqrt{5}+1\right)}{5-1}-\sqrt{27}+2\right)^{2}\left(2+\sqrt{3}\right)^{2}$$

Subtract 1 from 5 to get 4.

$$\left(-\left(3+\sqrt{6}\right)\left(\sqrt{2}-\sqrt{3}\right)+\frac{\left(\sqrt{15}-\sqrt{3}\right)\left(\sqrt{5}+1\right)}{4}-\sqrt{27}+2\right)^{2}\left(2+\sqrt{3}\right)^{2}$$

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}.

$$\left(-\left(3+\sqrt{6}\right)\left(\sqrt{2}-\sqrt{3}\right)+\frac{\left(\sqrt{15}-\sqrt{3}\right)\left(\sqrt{5}+1\right)}{4}-3\sqrt{3}+2\right)^{2}\left(2+\sqrt{3}\right)^{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(2+\sqrt{3}\right)^{2}.

$$\left(-\left(3+\sqrt{6}\right)\left(\sqrt{2}-\sqrt{3}\right)+\frac{\left(\sqrt{15}-\sqrt{3}\right)\left(\sqrt{5}+1\right)}{4}-3\sqrt{3}+2\right)^{2}\left(4+4\sqrt{3}+\left(\sqrt{3}\right)^{2}\right)$$

The square of \sqrt{3} is 3.

$$\left(-\left(3+\sqrt{6}\right)\left(\sqrt{2}-\sqrt{3}\right)+\frac{\left(\sqrt{15}-\sqrt{3}\right)\left(\sqrt{5}+1\right)}{4}-3\sqrt{3}+2\right)^{2}\left(4+4\sqrt{3}+3\right)$$

Add 4 and 3 to get 7.

$$\left(-\left(3+\sqrt{6}\right)\left(\sqrt{2}-\sqrt{3}\right)+\frac{\left(\sqrt{15}-\sqrt{3}\right)\left(\sqrt{5}+1\right)}{4}-3\sqrt{3}+2\right)^{2}\left(7+4\sqrt{3}\right)$$

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

$$\left(-\left(3\sqrt{2}-3\sqrt{3}+\sqrt{6}\sqrt{2}-\sqrt{6}\sqrt{3}\right)+\frac{\left(\sqrt{15}-\sqrt{3}\right)\left(\sqrt{5}+1\right)}{4}-3\sqrt{3}+2\right)^{2}\left(7+4\sqrt{3}\right)$$

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}.

$$\left(-\left(3\sqrt{2}-3\sqrt{3}+\sqrt{2}\sqrt{3}\sqrt{2}-\sqrt{6}\sqrt{3}\right)+\frac{\left(\sqrt{15}-\sqrt{3}\right)\left(\sqrt{5}+1\right)}{4}-3\sqrt{3}+2\right)^{2}\left(7+4\sqrt{3}\right)$$

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

$$\left(-\left(3\sqrt{2}-3\sqrt{3}+2\sqrt{3}-\sqrt{6}\sqrt{3}\right)+\frac{\left(\sqrt{15}-\sqrt{3}\right)\left(\sqrt{5}+1\right)}{4}-3\sqrt{3}+2\right)^{2}\left(7+4\sqrt{3}\right)$$

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

$$\left(-\left(3\sqrt{2}-\sqrt{3}-\sqrt{6}\sqrt{3}\right)+\frac{\left(\sqrt{15}-\sqrt{3}\right)\left(\sqrt{5}+1\right)}{4}-3\sqrt{3}+2\right)^{2}\left(7+4\sqrt{3}\right)$$

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}.

$$\left(-\left(3\sqrt{2}-\sqrt{3}-\sqrt{3}\sqrt{2}\sqrt{3}\right)+\frac{\left(\sqrt{15}-\sqrt{3}\right)\left(\sqrt{5}+1\right)}{4}-3\sqrt{3}+2\right)^{2}\left(7+4\sqrt{3}\right)$$

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

$$\left(-\left(3\sqrt{2}-\sqrt{3}-3\sqrt{2}\right)+\frac{\left(\sqrt{15}-\sqrt{3}\right)\left(\sqrt{5}+1\right)}{4}-3\sqrt{3}+2\right)^{2}\left(7+4\sqrt{3}\right)$$

Combine 3\sqrt{2} and -3\sqrt{2} to get 0.

$$\left(-\left(-\sqrt{3}\right)+\frac{\left(\sqrt{15}-\sqrt{3}\right)\left(\sqrt{5}+1\right)}{4}-3\sqrt{3}+2\right)^{2}\left(7+4\sqrt{3}\right)$$

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

$$\left(\sqrt{3}+\frac{\left(\sqrt{15}-\sqrt{3}\right)\left(\sqrt{5}+1\right)}{4}-3\sqrt{3}+2\right)^{2}\left(7+4\sqrt{3}\right)$$

Use the distributive property to multiply \sqrt{15}-\sqrt{3} by \sqrt{5}+1.

$$\left(\sqrt{3}+\frac{\sqrt{15}\sqrt{5}+\sqrt{15}-\sqrt{3}\sqrt{5}-\sqrt{3}}{4}-3\sqrt{3}+2\right)^{2}\left(7+4\sqrt{3}\right)$$

Factor 15=5\times 3. Rewrite the square root of the product \sqrt{5\times 3} as the product of square roots \sqrt{5}\sqrt{3}.

$$\left(\sqrt{3}+\frac{\sqrt{5}\sqrt{3}\sqrt{5}+\sqrt{15}-\sqrt{3}\sqrt{5}-\sqrt{3}}{4}-3\sqrt{3}+2\right)^{2}\left(7+4\sqrt{3}\right)$$

Multiply \sqrt{5} and \sqrt{5} to get 5.

$$\left(\sqrt{3}+\frac{5\sqrt{3}+\sqrt{15}-\sqrt{3}\sqrt{5}-\sqrt{3}}{4}-3\sqrt{3}+2\right)^{2}\left(7+4\sqrt{3}\right)$$

To multiply \sqrt{3} and \sqrt{5}, multiply the numbers under the square root.

$$\left(\sqrt{3}+\frac{5\sqrt{3}+\sqrt{15}-\sqrt{15}-\sqrt{3}}{4}-3\sqrt{3}+2\right)^{2}\left(7+4\sqrt{3}\right)$$

Combine \sqrt{15} and -\sqrt{15} to get 0.

$$\left(\sqrt{3}+\frac{5\sqrt{3}-\sqrt{3}}{4}-3\sqrt{3}+2\right)^{2}\left(7+4\sqrt{3}\right)$$

Combine 5\sqrt{3} and -\sqrt{3} to get 4\sqrt{3}.

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

Cancel out 4 and 4.

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

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

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

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

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

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(-\sqrt{3}+2\right)^{2}.

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

The square of \sqrt{3} is 3.

$$\left(3-4\sqrt{3}+4\right)\left(7+4\sqrt{3}\right)$$

Add 3 and 4 to get 7.

$$\left(7-4\sqrt{3}\right)\left(7+4\sqrt{3}\right)$$

Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}. Square 7.

$$49-\left(4\sqrt{3}\right)^{2}$$

Expand \left(4\sqrt{3}\right)^{2}.

$$49-4^{2}\left(\sqrt{3}\right)^{2}$$

Calculate 4 to the power of 2 and get 16.

$$49-16\left(\sqrt{3}\right)^{2}$$

The square of \sqrt{3} is 3.

$$49-16\times 3$$

Multiply 16 and 3 to get 48.

$$49-48$$

Subtract 48 from 49 to get 1.

$$1$$

Factor

$$1$$