Solve (frac{sqrt{5}+sqrt{3}}{sqrt{5}-sqrt{3}}+frac{sqrt{5}-sqrt{3}}{sqrt{5}+sqrt{3}})^2-(frac{sqrt{5}+sqrt{3}}{sqrt{5}-sqrt{3}}-frac{sqrt{5}-sqrt{3}}{sqrt{5}+sqrt{3}})=

$$( \frac { \sqrt { 5 } + \sqrt { 3 } } { \sqrt { 5 } - \sqrt { 3 } } + \frac { \sqrt { 5 } - \sqrt { 3 } } { \sqrt { 5 } + \sqrt { 3 } } ) ^ { 2 } - ( \frac { \sqrt { 5 } + \sqrt { 3 } } { \sqrt { 5 } - \sqrt { 3 } } - \frac { \sqrt { 5 } - \sqrt { 3 } } { \sqrt { 5 } + \sqrt { 3 } } ) =$$

Evaluate

$$64-2\sqrt{15}\approx 56.254033308$$

Solution Steps

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

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

Consider \left(\sqrt{5}-\sqrt{3}\right)\left(\sqrt{5}+\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}.

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

Square \sqrt{5}. Square \sqrt{3}.

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

Subtract 3 from 5 to get 2.

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

Multiply \sqrt{5}+\sqrt{3} and \sqrt{5}+\sqrt{3} to get \left(\sqrt{5}+\sqrt{3}\right)^{2}.

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

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(\sqrt{5}+\sqrt{3}\right)^{2}.

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

The square of \sqrt{5} is 5.

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

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

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

The square of \sqrt{3} is 3.

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

Add 5 and 3 to get 8.

$$\left(\frac{8+2\sqrt{15}}{2}+\frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{3}}\right)^{2}-\left(\frac{\sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}}-\frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{3}}\right)$$

Divide each term of 8+2\sqrt{15} by 2 to get 4+\sqrt{15}.

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

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

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

Consider \left(\sqrt{5}+\sqrt{3}\right)\left(\sqrt{5}-\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}.

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

Square \sqrt{5}. Square \sqrt{3}.

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

Subtract 3 from 5 to get 2.

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

Multiply \sqrt{5}-\sqrt{3} and \sqrt{5}-\sqrt{3} to get \left(\sqrt{5}-\sqrt{3}\right)^{2}.

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

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(\sqrt{5}-\sqrt{3}\right)^{2}.

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

The square of \sqrt{5} is 5.

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

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

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

The square of \sqrt{3} is 3.

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

Add 5 and 3 to get 8.

$$\left(4+\sqrt{15}+\frac{8-2\sqrt{15}}{2}\right)^{2}-\left(\frac{\sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}}-\frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{3}}\right)$$

Divide each term of 8-2\sqrt{15} by 2 to get 4-\sqrt{15}.

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

Add 4 and 4 to get 8.

$$\left(8+\sqrt{15}-\sqrt{15}\right)^{2}-\left(\frac{\sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}}-\frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{3}}\right)$$

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

$$8^{2}-\left(\frac{\sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}}-\frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{3}}\right)$$

Calculate 8 to the power of 2 and get 64.

$$64-\left(\frac{\sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}}-\frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{3}}\right)$$

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

$$64-\left(\frac{\left(\sqrt{5}+\sqrt{3}\right)\left(\sqrt{5}+\sqrt{3}\right)}{\left(\sqrt{5}-\sqrt{3}\right)\left(\sqrt{5}+\sqrt{3}\right)}-\frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{3}}\right)$$

Consider \left(\sqrt{5}-\sqrt{3}\right)\left(\sqrt{5}+\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}.

$$64-\left(\frac{\left(\sqrt{5}+\sqrt{3}\right)\left(\sqrt{5}+\sqrt{3}\right)}{\left(\sqrt{5}\right)^{2}-\left(\sqrt{3}\right)^{2}}-\frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{3}}\right)$$

Square \sqrt{5}. Square \sqrt{3}.

$$64-\left(\frac{\left(\sqrt{5}+\sqrt{3}\right)\left(\sqrt{5}+\sqrt{3}\right)}{5-3}-\frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{3}}\right)$$

Subtract 3 from 5 to get 2.

$$64-\left(\frac{\left(\sqrt{5}+\sqrt{3}\right)\left(\sqrt{5}+\sqrt{3}\right)}{2}-\frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{3}}\right)$$

Multiply \sqrt{5}+\sqrt{3} and \sqrt{5}+\sqrt{3} to get \left(\sqrt{5}+\sqrt{3}\right)^{2}.

$$64-\left(\frac{\left(\sqrt{5}+\sqrt{3}\right)^{2}}{2}-\frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{3}}\right)$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(\sqrt{5}+\sqrt{3}\right)^{2}.

$$64-\left(\frac{\left(\sqrt{5}\right)^{2}+2\sqrt{5}\sqrt{3}+\left(\sqrt{3}\right)^{2}}{2}-\frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{3}}\right)$$

The square of \sqrt{5} is 5.

$$64-\left(\frac{5+2\sqrt{5}\sqrt{3}+\left(\sqrt{3}\right)^{2}}{2}-\frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{3}}\right)$$

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

$$64-\left(\frac{5+2\sqrt{15}+\left(\sqrt{3}\right)^{2}}{2}-\frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{3}}\right)$$

The square of \sqrt{3} is 3.

$$64-\left(\frac{5+2\sqrt{15}+3}{2}-\frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{3}}\right)$$

Add 5 and 3 to get 8.

$$64-\left(\frac{8+2\sqrt{15}}{2}-\frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{3}}\right)$$

Divide each term of 8+2\sqrt{15} by 2 to get 4+\sqrt{15}.

$$64-\left(4+\sqrt{15}-\frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{3}}\right)$$

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

$$64-\left(4+\sqrt{15}-\frac{\left(\sqrt{5}-\sqrt{3}\right)\left(\sqrt{5}-\sqrt{3}\right)}{\left(\sqrt{5}+\sqrt{3}\right)\left(\sqrt{5}-\sqrt{3}\right)}\right)$$

Consider \left(\sqrt{5}+\sqrt{3}\right)\left(\sqrt{5}-\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}.

$$64-\left(4+\sqrt{15}-\frac{\left(\sqrt{5}-\sqrt{3}\right)\left(\sqrt{5}-\sqrt{3}\right)}{\left(\sqrt{5}\right)^{2}-\left(\sqrt{3}\right)^{2}}\right)$$

Square \sqrt{5}. Square \sqrt{3}.

$$64-\left(4+\sqrt{15}-\frac{\left(\sqrt{5}-\sqrt{3}\right)\left(\sqrt{5}-\sqrt{3}\right)}{5-3}\right)$$

Subtract 3 from 5 to get 2.

$$64-\left(4+\sqrt{15}-\frac{\left(\sqrt{5}-\sqrt{3}\right)\left(\sqrt{5}-\sqrt{3}\right)}{2}\right)$$

Multiply \sqrt{5}-\sqrt{3} and \sqrt{5}-\sqrt{3} to get \left(\sqrt{5}-\sqrt{3}\right)^{2}.

$$64-\left(4+\sqrt{15}-\frac{\left(\sqrt{5}-\sqrt{3}\right)^{2}}{2}\right)$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(\sqrt{5}-\sqrt{3}\right)^{2}.

$$64-\left(4+\sqrt{15}-\frac{\left(\sqrt{5}\right)^{2}-2\sqrt{5}\sqrt{3}+\left(\sqrt{3}\right)^{2}}{2}\right)$$

The square of \sqrt{5} is 5.

$$64-\left(4+\sqrt{15}-\frac{5-2\sqrt{5}\sqrt{3}+\left(\sqrt{3}\right)^{2}}{2}\right)$$

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

$$64-\left(4+\sqrt{15}-\frac{5-2\sqrt{15}+\left(\sqrt{3}\right)^{2}}{2}\right)$$

The square of \sqrt{3} is 3.

$$64-\left(4+\sqrt{15}-\frac{5-2\sqrt{15}+3}{2}\right)$$

Add 5 and 3 to get 8.

$$64-\left(4+\sqrt{15}-\frac{8-2\sqrt{15}}{2}\right)$$

Divide each term of 8-2\sqrt{15} by 2 to get 4-\sqrt{15}.

$$64-\left(4+\sqrt{15}-\left(4-\sqrt{15}\right)\right)$$

To find the opposite of 4-\sqrt{15}, find the opposite of each term.

$$64-\left(4+\sqrt{15}-4+\sqrt{15}\right)$$

Subtract 4 from 4 to get 0.

$$64-\left(\sqrt{15}+\sqrt{15}\right)$$

Combine \sqrt{15} and \sqrt{15} to get 2\sqrt{15}.

$$64-2\sqrt{15}$$