Solve frac{(6+2sqrt{3})^2+(4sqrt{3})^2-(2sqrt{6})^2}{2*(6+2sqrt{3})*(4sqrt{3})}

$$\frac { ( 6 + 2 \sqrt { 3 } ) ^ { 2 } + ( 4 \sqrt { 3 } ) ^ { 2 } - ( 2 \sqrt { 6 } ) ^ { 2 } } { 2 \times ( 6 + 2 \sqrt { 3 } ) \times ( 4 \sqrt { 3 } ) }$$

Evaluate

$$\frac{\sqrt{3}}{2}\approx 0.866025404$$

Solution Steps

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(6+2\sqrt{3}\right)^{2}.

$$\frac{36+24\sqrt{3}+4\left(\sqrt{3}\right)^{2}+\left(4\sqrt{3}\right)^{2}-\left(2\sqrt{6}\right)^{2}}{2\left(6+2\sqrt{3}\right)\times 4\sqrt{3}}$$

The square of \sqrt{3} is 3.

$$\frac{36+24\sqrt{3}+4\times 3+\left(4\sqrt{3}\right)^{2}-\left(2\sqrt{6}\right)^{2}}{2\left(6+2\sqrt{3}\right)\times 4\sqrt{3}}$$

Multiply 4 and 3 to get 12.

$$\frac{36+24\sqrt{3}+12+\left(4\sqrt{3}\right)^{2}-\left(2\sqrt{6}\right)^{2}}{2\left(6+2\sqrt{3}\right)\times 4\sqrt{3}}$$

Add 36 and 12 to get 48.

$$\frac{48+24\sqrt{3}+\left(4\sqrt{3}\right)^{2}-\left(2\sqrt{6}\right)^{2}}{2\left(6+2\sqrt{3}\right)\times 4\sqrt{3}}$$

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

$$\frac{48+24\sqrt{3}+4^{2}\left(\sqrt{3}\right)^{2}-\left(2\sqrt{6}\right)^{2}}{2\left(6+2\sqrt{3}\right)\times 4\sqrt{3}}$$

Calculate 4 to the power of 2 and get 16.

$$\frac{48+24\sqrt{3}+16\left(\sqrt{3}\right)^{2}-\left(2\sqrt{6}\right)^{2}}{2\left(6+2\sqrt{3}\right)\times 4\sqrt{3}}$$

The square of \sqrt{3} is 3.

$$\frac{48+24\sqrt{3}+16\times 3-\left(2\sqrt{6}\right)^{2}}{2\left(6+2\sqrt{3}\right)\times 4\sqrt{3}}$$

Multiply 16 and 3 to get 48.

$$\frac{48+24\sqrt{3}+48-\left(2\sqrt{6}\right)^{2}}{2\left(6+2\sqrt{3}\right)\times 4\sqrt{3}}$$

Add 48 and 48 to get 96.

$$\frac{96+24\sqrt{3}-\left(2\sqrt{6}\right)^{2}}{2\left(6+2\sqrt{3}\right)\times 4\sqrt{3}}$$

Expand \left(2\sqrt{6}\right)^{2}.

$$\frac{96+24\sqrt{3}-2^{2}\left(\sqrt{6}\right)^{2}}{2\left(6+2\sqrt{3}\right)\times 4\sqrt{3}}$$

Calculate 2 to the power of 2 and get 4.

$$\frac{96+24\sqrt{3}-4\left(\sqrt{6}\right)^{2}}{2\left(6+2\sqrt{3}\right)\times 4\sqrt{3}}$$

The square of \sqrt{6} is 6.

$$\frac{96+24\sqrt{3}-4\times 6}{2\left(6+2\sqrt{3}\right)\times 4\sqrt{3}}$$

Multiply 4 and 6 to get 24.

$$\frac{96+24\sqrt{3}-24}{2\left(6+2\sqrt{3}\right)\times 4\sqrt{3}}$$

Subtract 24 from 96 to get 72.

$$\frac{72+24\sqrt{3}}{2\left(6+2\sqrt{3}\right)\times 4\sqrt{3}}$$

Multiply 2 and 4 to get 8.

$$\frac{72+24\sqrt{3}}{8\left(6+2\sqrt{3}\right)\sqrt{3}}$$

Rationalize the denominator of \frac{72+24\sqrt{3}}{8\left(6+2\sqrt{3}\right)\sqrt{3}} by multiplying numerator and denominator by \sqrt{3}.

$$\frac{\left(72+24\sqrt{3}\right)\sqrt{3}}{8\left(6+2\sqrt{3}\right)\left(\sqrt{3}\right)^{2}}$$

The square of \sqrt{3} is 3.

$$\frac{\left(72+24\sqrt{3}\right)\sqrt{3}}{8\left(6+2\sqrt{3}\right)\times 3}$$

Multiply 8 and 3 to get 24.

$$\frac{\left(72+24\sqrt{3}\right)\sqrt{3}}{24\left(6+2\sqrt{3}\right)}$$

Use the distributive property to multiply 72+24\sqrt{3} by \sqrt{3}.

$$\frac{72\sqrt{3}+24\left(\sqrt{3}\right)^{2}}{24\left(6+2\sqrt{3}\right)}$$

The square of \sqrt{3} is 3.

$$\frac{72\sqrt{3}+24\times 3}{24\left(6+2\sqrt{3}\right)}$$

Multiply 24 and 3 to get 72.

$$\frac{72\sqrt{3}+72}{24\left(6+2\sqrt{3}\right)}$$

Use the distributive property to multiply 24 by 6+2\sqrt{3}.

$$\frac{72\sqrt{3}+72}{144+48\sqrt{3}}$$

Rationalize the denominator of \frac{72\sqrt{3}+72}{144+48\sqrt{3}} by multiplying numerator and denominator by 144-48\sqrt{3}.

$$\frac{\left(72\sqrt{3}+72\right)\left(144-48\sqrt{3}\right)}{\left(144+48\sqrt{3}\right)\left(144-48\sqrt{3}\right)}$$

Consider \left(144+48\sqrt{3}\right)\left(144-48\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}.

$$\frac{\left(72\sqrt{3}+72\right)\left(144-48\sqrt{3}\right)}{144^{2}-\left(48\sqrt{3}\right)^{2}}$$

Calculate 144 to the power of 2 and get 20736.

$$\frac{\left(72\sqrt{3}+72\right)\left(144-48\sqrt{3}\right)}{20736-\left(48\sqrt{3}\right)^{2}}$$

Expand \left(48\sqrt{3}\right)^{2}.

$$\frac{\left(72\sqrt{3}+72\right)\left(144-48\sqrt{3}\right)}{20736-48^{2}\left(\sqrt{3}\right)^{2}}$$

Calculate 48 to the power of 2 and get 2304.

$$\frac{\left(72\sqrt{3}+72\right)\left(144-48\sqrt{3}\right)}{20736-2304\left(\sqrt{3}\right)^{2}}$$

The square of \sqrt{3} is 3.

$$\frac{\left(72\sqrt{3}+72\right)\left(144-48\sqrt{3}\right)}{20736-2304\times 3}$$

Multiply 2304 and 3 to get 6912.

$$\frac{\left(72\sqrt{3}+72\right)\left(144-48\sqrt{3}\right)}{20736-6912}$$

Subtract 6912 from 20736 to get 13824.

$$\frac{\left(72\sqrt{3}+72\right)\left(144-48\sqrt{3}\right)}{13824}$$

Use the distributive property to multiply 72\sqrt{3}+72 by 144-48\sqrt{3} and combine like terms.

$$\frac{6912\sqrt{3}-3456\left(\sqrt{3}\right)^{2}+10368}{13824}$$

The square of \sqrt{3} is 3.

$$\frac{6912\sqrt{3}-3456\times 3+10368}{13824}$$

Multiply -3456 and 3 to get -10368.

$$\frac{6912\sqrt{3}-10368+10368}{13824}$$

Add -10368 and 10368 to get 0.

$$\frac{6912\sqrt{3}}{13824}$$

Divide 6912\sqrt{3} by 13824 to get \frac{1}{2}\sqrt{3}.

$$\frac{1}{2}\sqrt{3}$$