Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)

Published by Pearson
ISBN 10: 0-32193-104-1
ISBN 13: 978-0-32193-104-7

Appendix A - Review - A.7 nth Roots; Rational Exponents - A.7 Assess Your Understanding - Page A61: 63


$\dfrac{27}{16\sqrt 2}$

Work Step by Step

Since $9=3^2$ and $8=2(2^2)$, we have $$ \left(\frac{9}{8}\right)^{3/2}=\left(\frac{3^2}{2(2^2)}\right)^{3/2}=\frac{3^3}{2^3\sqrt{2^3}}=\frac{27}{16\sqrt 2} .$$
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