## Algebra and Trigonometry 10th Edition

(a) $2\sqrt 5$ (b) $4\sqrt[3] 2$
(a) $\sqrt {20}=\sqrt {4\times5}=\sqrt {2^2\times5}=2\sqrt 5$ (b) $\sqrt[3] {128}=\sqrt[3] {64\times2}=\sqrt[3] {2^6\times2}=2^2\sqrt[3] 2=4\sqrt[3] 2$