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 - Page 46: 82

Answer

$\sqrt{3}+\sqrt[3]{15}$

Work Step by Step

The radicand of $\sqrt[3]{15}$ has no perfect cube factor. This means that the radical is already in its simplest form.
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.