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

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

Evaluate

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

Solution Steps

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

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

Subtract 1 from 1 to get 0.

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

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

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

Calculate 2 to the power of 2 and get 4.

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

The square of \sqrt{3} is 3.

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

Multiply 4 and 3 to get 12.

$$\frac{12}{\left(\sqrt{3}+1\right)^{2}-\left(\sqrt{3}-1\right)^{2}}$$

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

$$\frac{12}{\left(\sqrt{3}\right)^{2}+2\sqrt{3}+1-\left(\sqrt{3}-1\right)^{2}}$$

The square of \sqrt{3} is 3.

$$\frac{12}{3+2\sqrt{3}+1-\left(\sqrt{3}-1\right)^{2}}$$

Add 3 and 1 to get 4.

$$\frac{12}{4+2\sqrt{3}-\left(\sqrt{3}-1\right)^{2}}$$

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

$$\frac{12}{4+2\sqrt{3}-\left(\left(\sqrt{3}\right)^{2}-2\sqrt{3}+1\right)}$$

The square of \sqrt{3} is 3.

$$\frac{12}{4+2\sqrt{3}-\left(3-2\sqrt{3}+1\right)}$$

Add 3 and 1 to get 4.

$$\frac{12}{4+2\sqrt{3}-\left(4-2\sqrt{3}\right)}$$

To find the opposite of 4-2\sqrt{3}, find the opposite of each term.

$$\frac{12}{4+2\sqrt{3}-4+2\sqrt{3}}$$

Subtract 4 from 4 to get 0.

$$\frac{12}{2\sqrt{3}+2\sqrt{3}}$$

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

$$\frac{12}{4\sqrt{3}}$$

Rationalize the denominator of \frac{12}{4\sqrt{3}} by multiplying numerator and denominator by \sqrt{3}.

$$\frac{12\sqrt{3}}{4\left(\sqrt{3}\right)^{2}}$$

The square of \sqrt{3} is 3.

$$\frac{12\sqrt{3}}{4\times 3}$$

Cancel out 3\times 4 in both numerator and denominator.

$$\sqrt{3}$$

Expand

$$\sqrt{3} = 1.732050808$$

Solution Steps

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

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

Subtract 1 from 1 to get 0.

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

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

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

Calculate 2 to the power of 2 and get 4.

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

The square of \sqrt{3} is 3.

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

Multiply 4 and 3 to get 12.

$$\frac{12}{\left(\sqrt{3}+1\right)^{2}-\left(\sqrt{3}-1\right)^{2}}$$

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

$$\frac{12}{\left(\sqrt{3}\right)^{2}+2\sqrt{3}+1-\left(\sqrt{3}-1\right)^{2}}$$

The square of \sqrt{3} is 3.

$$\frac{12}{3+2\sqrt{3}+1-\left(\sqrt{3}-1\right)^{2}}$$

Add 3 and 1 to get 4.

$$\frac{12}{4+2\sqrt{3}-\left(\sqrt{3}-1\right)^{2}}$$

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

$$\frac{12}{4+2\sqrt{3}-\left(\left(\sqrt{3}\right)^{2}-2\sqrt{3}+1\right)}$$

The square of \sqrt{3} is 3.

$$\frac{12}{4+2\sqrt{3}-\left(3-2\sqrt{3}+1\right)}$$

Add 3 and 1 to get 4.

$$\frac{12}{4+2\sqrt{3}-\left(4-2\sqrt{3}\right)}$$

To find the opposite of 4-2\sqrt{3}, find the opposite of each term.

$$\frac{12}{4+2\sqrt{3}-4+2\sqrt{3}}$$

Subtract 4 from 4 to get 0.

$$\frac{12}{2\sqrt{3}+2\sqrt{3}}$$

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

$$\frac{12}{4\sqrt{3}}$$

Rationalize the denominator of \frac{12}{4\sqrt{3}} by multiplying numerator and denominator by \sqrt{3}.

$$\frac{12\sqrt{3}}{4\left(\sqrt{3}\right)^{2}}$$

The square of \sqrt{3} is 3.

$$\frac{12\sqrt{3}}{4\times 3}$$

Cancel out 3\times 4 in both numerator and denominator.

$$\sqrt{3}$$