Evaluate: $\left( \frac{1}{2} + \frac{1}{3} \right) \left( \frac{1}{2} - \frac{1}{3} \right)$
I notice that this expression is of the form $(a + b)(a - b)$, which is a difference of squares. I recall that $(a + b)(a - b) = a^2 - b^2$, so I can apply that formula to simplify the expression. I get: $\left( \frac{1}{2} + \frac{1}{3} \right) \left( \frac{1}{2} - \frac{1}{3} \right) = \left( \frac{1}{2} \right)^2 - \left( \frac{1}{3} \right)^2$ Now I just need to evaluate the squares of the fractions. I know that $\left( \frac{1}{2} \right)^2 = \frac{1}{4}$ and $\left( \frac{1}{3} \right)^2 = \frac{1}{9}$. So I substitute those values and get: $\left( \frac{1}{2} \right)^2 - \left( \frac{1}{3} \right)^2 = \frac{1}{4} - \frac{1}{9}$ To subtract the fractions, I need to find a common denominator. The least common multiple of 4 and 9 is 36, so I multiply the first fraction by $\frac{9}{9}$ and the second fraction by $\frac{4}{4}$. I get: $\frac{1}{4} - \frac{1}{9} = \frac{9}{36} - \frac{4}{36} = \frac{9 - 4}{36} = \frac{5}{36}$ 