Intermediate Algebra (12th Edition)

Published by Pearson
ISBN 10: 0321969359
ISBN 13: 978-0-32196-935-4

Chapter 7 - Section 7.3 - Simplifying Radicals, the Distance Formula, and Circles - 7.3 Exercises - Page 460: 108



Work Step by Step

$\bf{\text{Solution Outline:}}$ To simplify the given expression, $ \sqrt[3]{5}\cdot\sqrt{6} ,$ express the radicals as radicals with same indices by finding the $LCD$ of the indices. Once the indices are the same, use the laws of radicals to simplify the expression. $\bf{\text{Solution Details:}}$ The $LCD$ of the indices, $ 3 $ and $ 2 ,$ is $ 6 $ since it is the lowest number that can be divided exactly by both indices. Multiplying the index by a number to make it equal to the $LCD$ and raising the radicand by the same multiplier results to \begin{array}{l}\require{cancel} \sqrt[3(2)]{5^2}\cdot\sqrt[2(3)]{6^3} \\\\= \sqrt[6]{25}\cdot\sqrt[6]{216} .\end{array} Using the Product Rule of radicals which is given by $\sqrt[m]{x}\cdot\sqrt[m]{y}=\sqrt[m]{xy},$ the expression above is equivalent to\begin{array}{l}\require{cancel} \sqrt[6]{25(216)} \\\\= \sqrt[6]{5,400} .\end{array}
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