## Precalculus: Mathematics for Calculus, 7th Edition

a) $3$$\sqrt5 b) 4 c) 6$$\sqrt[3] 2$
a) $\sqrt 3$$\sqrt 15 = \sqrt {3\times3\times5} = \sqrt 9$$\sqrt 5$ = 3$\sqrt 5$ b) $\frac{\sqrt {48}}{\sqrt 3}$ = $\frac{\sqrt {3\times2\times8}}{\sqrt 3}$ = $\sqrt {16}$ = 4 c) $\sqrt[3]{24}$$\sqrt[3] {18}= \sqrt[3] {6^{3}\times2} = 6$$\sqrt[3] 2$