Intermediate Algebra (6th Edition)

Published by Pearson
ISBN 10: 0321785045
ISBN 13: 978-0-32178-504-6

Chapter 9 - Cumulative Review - Page 601: 28c

Answer

$9ab^{2}\sqrt[3] {a^{2}}$

Work Step by Step

$\frac{3\sqrt[3] {81a^{5}b^{10}}}{\sqrt[3] {3b^{4}}}$ =$\frac{3\sqrt[3] {27\times3\times a^{3} \times a^{2} \times b^{9}\times b}}{\sqrt[3] {3b\times b^{3}}}$ =$\frac{3\sqrt[3] {27\times a^{3} \times b^{9}\times 3 \times a^{2} \times b}}{b\sqrt[3] {3b}}$ =$\frac{3\sqrt[3] {27\times a^{3} \times b^{9}}\times \sqrt[3] {3 \times a^{2} \times b}}{b\sqrt[3] {3b}}$ =$\frac{3\times 3ab^{3}\times \sqrt[3] {3 \times a^{2} \times b}}{b\sqrt[3] {3b}}$ =$\frac{3\times 3ab^{2}\times \sqrt[3] {3 \times a^{2} \times b}}{\sqrt[3] {3b}}$ =$\frac{9ab^{2}\times \sqrt[3] {3b}\times \sqrt[3] {a^{2}}}{\sqrt[3] {3b}}$ =$9ab^{2}\sqrt[3] {a^{2}}$
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.