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

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

Evaluate

$$\frac{8}{3}\approx 2.666666667$$

Short Solution Steps

To add or subtract expressions, expand them to make their denominators the same. Multiply 1 times \frac{3}{3}.

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

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

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

To raise \frac{3+\sqrt{3}}{3} to a power, raise both numerator and denominator to the power and then divide.

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

To add or subtract expressions, expand them to make their denominators the same. Multiply 1 times \frac{3}{3}.

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

Since \frac{3}{3} and \frac{\sqrt{3}}{3} have the same denominator, subtract them by subtracting their numerators.

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

To raise \frac{3-\sqrt{3}}{3} to a power, raise both numerator and denominator to the power and then divide.

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

Since \frac{\left(3+\sqrt{3}\right)^{2}}{3^{2}} and \frac{\left(3-\sqrt{3}\right)^{2}}{3^{2}} have the same denominator, add them by adding their numerators.

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

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

$$\frac{9+6\sqrt{3}+\left(\sqrt{3}\right)^{2}}{3^{2}}+\frac{\left(3-\sqrt{3}\right)^{2}}{3^{2}}$$

The square of \sqrt{3} is 3.

$$\frac{9+6\sqrt{3}+3}{3^{2}}+\frac{\left(3-\sqrt{3}\right)^{2}}{3^{2}}$$

Add 9 and 3 to get 12.

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

Calculate 3 to the power of 2 and get 9.

$$\frac{12+6\sqrt{3}}{9}+\frac{\left(3-\sqrt{3}\right)^{2}}{3^{2}}$$

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

$$\frac{12+6\sqrt{3}}{9}+\frac{9-6\sqrt{3}+\left(\sqrt{3}\right)^{2}}{3^{2}}$$

The square of \sqrt{3} is 3.

$$\frac{12+6\sqrt{3}}{9}+\frac{9-6\sqrt{3}+3}{3^{2}}$$

Add 9 and 3 to get 12.

$$\frac{12+6\sqrt{3}}{9}+\frac{12-6\sqrt{3}}{3^{2}}$$

Calculate 3 to the power of 2 and get 9.

$$\frac{12+6\sqrt{3}}{9}+\frac{12-6\sqrt{3}}{9}$$

Since \frac{12+6\sqrt{3}}{9} and \frac{12-6\sqrt{3}}{9} have the same denominator, add them by adding their numerators.

$$\frac{12+6\sqrt{3}+12-6\sqrt{3}}{9}$$

Do the calculations in 12+6\sqrt{3}+12-6\sqrt{3}.

$$\frac{24}{9}$$

Factor

$$\frac{2 ^ {3}}{3} = 2\frac{2}{3} = 2.6666666666666665$$