## Precalculus (6th Edition) Blitzer

$3$
Recognize powers of 2 and 5, $\qquad 16=2^{4}=4^{2},\quad 625=5^{4}=25^{2}$ $\sqrt[3]{{\sqrt[4]{16}}+{\sqrt[2]{625}}}=\sqrt[3]{2+25}$ $=\sqrt[3]{2+25}=\sqrt[3]{27}$ ... recognize a power of 3, $\quad 27=3^{3}$ $=3$