## Intermediate Algebra (6th Edition)

Published by Pearson

# Chapter 8 - Cumulative Review - Page 533: 6a

#### Answer

$\frac{a^{6}}{b^{3}c^{9}}$

#### Work Step by Step

$(a^{-2}bc^{3})^{-3}$ =$(a^{-2})^{-3}\times b^{-3} \times (c^{3})^{-3}$ =$(a^{6})\times b^{-3} \times (c^{-9})$ =$(a^{6})\times \frac{1}{b^{3}} \times \frac{1}{c^{9}}$ =$\frac{a^{6}}{b^{3}c^{9}}$

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