Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter P - Section P.3 - Radicals and Rational Exponents - Exercise Set - Page 46: 109

Answer

$3$

Work Step by Step

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, $\quad 27=3^{3}$ $=3$
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