Precalculus (6th Edition)

Published by Pearson
ISBN 10: 013421742X
ISBN 13: 978-0-13421-742-0

Chapter R - Review of Basic Concepts - R.6 Rational Exponents - R.6 Exercises: 52

Answer

$\color{blue}{-\dfrac{2}{3}}$

Work Step by Step

RECALL: (1) $a^{1/n} = \sqrt[n]{a}$ (2) If $a \gt 0$, then $\sqrt[n]{a^n} = a$ (3) If $n$ is odd, then $\sqrt[n]{a^n}=a$ Use rule (1) above to obtain: $\left(-\dfrac{8}{27}\right)^{1/3} = \sqrt[3]{-\left(\dfrac{8}{27}\right)}$ Since $-\dfrac{8}{27}=\left(-\dfrac{2}{3}\right)^3$, then the expression above is equivalent to: $=\sqrt[3]{\left(-\dfrac{2}{3}\right)^3}$ Use rule (3) above to simplify and have: $\sqrt[3]{\left(-\dfrac{2}{3}\right)^3}=\color{blue}{-\dfrac{2}{3}}$
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