Solve {left(frac{sqrt{6}-sqrt{2}}{4}right)}^2+{left(frac{sqrt{6}+sqrt{2}}{4}right)}^2

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

Evaluate

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Short Solution Steps

To raise \frac{\sqrt{6}-\sqrt{2}}{4} to a power, raise both numerator and denominator to the power and then divide.

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

To raise \frac{\sqrt{6}+\sqrt{2}}{4} to a power, raise both numerator and denominator to the power and then divide.

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

Since \frac{\left(\sqrt{6}-\sqrt{2}\right)^{2}}{4^{2}} and \frac{\left(\sqrt{6}+\sqrt{2}\right)^{2}}{4^{2}} have the same denominator, add them by adding their numerators.

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

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

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

The square of \sqrt{6} is 6.

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

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}.

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

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

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

Multiply -2 and 2 to get -4.

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

The square of \sqrt{2} is 2.

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

Add 6 and 2 to get 8.

$$\frac{8-4\sqrt{3}}{4^{2}}+\frac{\left(\sqrt{6}+\sqrt{2}\right)^{2}}{4^{2}}$$

Calculate 4 to the power of 2 and get 16.

$$\frac{8-4\sqrt{3}}{16}+\frac{\left(\sqrt{6}+\sqrt{2}\right)^{2}}{4^{2}}$$

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

$$\frac{8-4\sqrt{3}}{16}+\frac{\left(\sqrt{6}\right)^{2}+2\sqrt{6}\sqrt{2}+\left(\sqrt{2}\right)^{2}}{4^{2}}$$

The square of \sqrt{6} is 6.

$$\frac{8-4\sqrt{3}}{16}+\frac{6+2\sqrt{6}\sqrt{2}+\left(\sqrt{2}\right)^{2}}{4^{2}}$$

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}.

$$\frac{8-4\sqrt{3}}{16}+\frac{6+2\sqrt{2}\sqrt{3}\sqrt{2}+\left(\sqrt{2}\right)^{2}}{4^{2}}$$

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

$$\frac{8-4\sqrt{3}}{16}+\frac{6+2\times 2\sqrt{3}+\left(\sqrt{2}\right)^{2}}{4^{2}}$$

Multiply 2 and 2 to get 4.

$$\frac{8-4\sqrt{3}}{16}+\frac{6+4\sqrt{3}+\left(\sqrt{2}\right)^{2}}{4^{2}}$$

The square of \sqrt{2} is 2.

$$\frac{8-4\sqrt{3}}{16}+\frac{6+4\sqrt{3}+2}{4^{2}}$$

Add 6 and 2 to get 8.

$$\frac{8-4\sqrt{3}}{16}+\frac{8+4\sqrt{3}}{4^{2}}$$

Calculate 4 to the power of 2 and get 16.

$$\frac{8-4\sqrt{3}}{16}+\frac{8+4\sqrt{3}}{16}$$

Since \frac{8-4\sqrt{3}}{16} and \frac{8+4\sqrt{3}}{16} have the same denominator, add them by adding their numerators.

$$\frac{8-4\sqrt{3}+8+4\sqrt{3}}{16}$$

Do the calculations in 8-4\sqrt{3}+8+4\sqrt{3}.

$$\frac{16}{16}$$

Factor

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