Solve 1/3+sqrt{3}+1/1+sqrt{3}

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

Evaluate

$$\frac{\sqrt{3}}{3}\approx 0.577350269$$

Short Solution Steps

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

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

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

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

Square 3. Square \sqrt{3}.

$$\frac{3-\sqrt{3}}{9-3}+\frac{1}{1+\sqrt{3}}$$

Subtract 3 from 9 to get 6.

$$\frac{3-\sqrt{3}}{6}+\frac{1}{1+\sqrt{3}}$$

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

$$\frac{3-\sqrt{3}}{6}+\frac{1-\sqrt{3}}{\left(1+\sqrt{3}\right)\left(1-\sqrt{3}\right)}$$

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}.

$$\frac{3-\sqrt{3}}{6}+\frac{1-\sqrt{3}}{1^{2}-\left(\sqrt{3}\right)^{2}}$$

Square 1. Square \sqrt{3}.

$$\frac{3-\sqrt{3}}{6}+\frac{1-\sqrt{3}}{1-3}$$

Subtract 3 from 1 to get -2.

$$\frac{3-\sqrt{3}}{6}+\frac{1-\sqrt{3}}{-2}$$

Multiply both numerator and denominator by -1.

$$\frac{3-\sqrt{3}}{6}+\frac{-1+\sqrt{3}}{2}$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of 6 and 2 is 6. Multiply \frac{-1+\sqrt{3}}{2} times \frac{3}{3}.

$$\frac{3-\sqrt{3}}{6}+\frac{3\left(-1+\sqrt{3}\right)}{6}$$

Since \frac{3-\sqrt{3}}{6} and \frac{3\left(-1+\sqrt{3}\right)}{6} have the same denominator, add them by adding their numerators.

$$\frac{3-\sqrt{3}+3\left(-1+\sqrt{3}\right)}{6}$$

Do the multiplications in 3-\sqrt{3}+3\left(-1+\sqrt{3}\right).

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

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

$$\frac{2\sqrt{3}}{6}$$

Divide 2\sqrt{3} by 6 to get \frac{1}{3}\sqrt{3}.

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