Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter P - Section P.3 - Radicals and Rational Exponents - Exercise Set: 77

Answer

$13\sqrt[3]{2}$

Work Step by Step

Simplify each radicand to obtain $5\sqrt[3]{8(2)}+\sqrt[3]{27(2)} \\=5\sqrt[3]{2^3(2)}+\sqrt[3]{3^3(2)} \\=5\cdot2\sqrt[3]{2} + 3\sqrt [3]{2} \\=10\sqrt[3]{2} + 3\sqrt[3]{2}.$ RECALL: The distributive property states that for any real numbers a, b, and c, $ac + bc = (a+b)c.$ Use the distributive to obtain $(10+3)\sqrt[3]{2} \\=13\sqrt[3]{2}.$
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.