14 Is each of these numbers rational or irrational? Give a reason for each answer. \( \begin{array}{llll}\text { a } \sqrt{3}+6 & \text { b } \sqrt{3+6} & \text { c } \sqrt{64}+\sqrt[3]{64} & \text { d } \sqrt[3]{8}+\sqrt[3]{19}\end{array} \)

The number \( \sqrt{3} + 6 \) is irrational because \( \sqrt{3} \) itself is an irrational number, and the sum of an irrational number and a rational number (6) remains irrational. The number \( \sqrt{3+6} \) simplifies to \( \sqrt{9} \), which equals 3. Since 3 is a whole number, it is a rational number. The expression \( \sqrt{64} + \sqrt[3]{64} \) equals \( 8 + 4 \) (since \( \sqrt{64} = 8 \) and \( \sqrt[3]{64} = 4 \)). The result, 12, is a rational number. Finally, \( \sqrt[3]{8} + \sqrt[3]{19} \) yields \( 2 + \sqrt[3]{19} \). Since \( \sqrt[3]{19} \) is irrational, the sum remains irrational.