Solve {left(frac{1-sqrt{3}}{1+sqrt{3}}right)}^2+{left(frac{1+sqrt{3}}{1-sqrt{3}}right)}^2

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

Evaluate

$$14$$

Solution Steps

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

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

Consider \left(1+\sqrt{3}\right)\left(1-\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}.

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

Square 1. Square \sqrt{3}.

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

Subtract 3 from 1 to get -2.

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

Multiply 1-\sqrt{3} and 1-\sqrt{3} to get \left(1-\sqrt{3}\right)^{2}.

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

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(1-\sqrt{3}\right)^{2}.

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

The square of \sqrt{3} is 3.

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

Add 1 and 3 to get 4.

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

Divide each term of 4-2\sqrt{3} by -2 to get -2+\sqrt{3}.

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

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

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

The square of \sqrt{3} is 3.

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

Add 4 and 3 to get 7.

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

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

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

Consider \left(1-\sqrt{3}\right)\left(1+\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}.

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

Square 1. Square \sqrt{3}.

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

Subtract 3 from 1 to get -2.

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

Multiply 1+\sqrt{3} and 1+\sqrt{3} to get \left(1+\sqrt{3}\right)^{2}.

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

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(1+\sqrt{3}\right)^{2}.

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

The square of \sqrt{3} is 3.

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

Add 1 and 3 to get 4.

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

Divide each term of 4+2\sqrt{3} by -2 to get -2-\sqrt{3}.

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

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

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

The square of \sqrt{3} is 3.

$$7-4\sqrt{3}+4+4\sqrt{3}+3$$

Add 4 and 3 to get 7.

$$7-4\sqrt{3}+7+4\sqrt{3}$$

Add 7 and 7 to get 14.

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

Combine -4\sqrt{3} and 4\sqrt{3} to get 0.

$$14$$

Factor

$$2\times 7$$