Algebra: A Combined Approach (4th Edition)

Published by Pearson
ISBN 10: 0321726391
ISBN 13: 978-0-32172-639-1

Chapter 12 - Cumulative Review: 34

Answer

$\dfrac{3\sqrt [3] {m^2n}}{m^2n^3}$

Work Step by Step

$\sqrt[3] {\dfrac{27}{m^4n^8}} = \dfrac{\sqrt[3]{27}}{\sqrt[3] {m^4n^8}} = \dfrac{3}{\sqrt[3] {m^4n^8}} = \dfrac{3}{\sqrt[3] {m^4n^8}} \times \dfrac{\sqrt[3] {m^2n}}{\sqrt[3] {m^2n}} = \dfrac{3\sqrt [3] {m^2n}}{\sqrt[3] {m^4n^8} \sqrt [3] {m^2n}} = \dfrac{3\sqrt [3] {m^2n}}{\sqrt[3] {m^4n^8m^2n}} = \dfrac{3\sqrt [3] {m^2n}}{\sqrt[3] {m^{4+2}n^{8+1}}} = \dfrac{3\sqrt [3] {m^2n}}{\sqrt[3] {m^6n^9}} = \dfrac{3\sqrt [3] {m^2n}}{\sqrt[3] {(m^2)^3(n^3)^3}} = \dfrac{3\sqrt [3] {m^2n}}{m^2n^3}$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.