Chapter 2 - Real Numbers - Chapter 2 Review Problem Set - Page 88: 14

$= \frac{4}{9}$

Recall, because of PEDMAS, we consider what is inside of parenthesis first, and then we consider the exponent. Thus, we obtain: $= (\frac{1}{2} + \frac{1}{3} - \frac{1}{6})^{2}$ $=(\frac{3}{6} +\frac{2}{6}-\frac{1}{6})^{2}$ $= (\frac{5}{6} - \frac{1}{6})^{2}$ $= (\frac{5}{6} - \frac{1}{6})^{2}$ $= (\frac{4}{6})^{2}$ $= (\frac{2}{3})^{2}$ $= (\frac{4}{9})$