#### Answer

$\dfrac{1}{16}$

#### Work Step by Step

$64^{-(2/3)}=\dfrac{1}{64^{2/3}}=\dfrac{1}{\sqrt[3] {64^{2}}}=\dfrac{1}{\sqrt[3] {4096}}=\dfrac{1}{16}$

Published by
Pearson

ISBN 10:
0321726391

ISBN 13:
978-0-32172-639-1

$\dfrac{1}{16}$

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