College Algebra (11th Edition)

Published by Pearson
ISBN 10: 0321671791
ISBN 13: 978-0-32167-179-0

Chapter R - Section R.7 - Radical Expressions - R.7 Exercises: 50

Answer

$-15\sqrt[3]{-75}$

Work Step by Step

$\bf{\text{Solution Outline:}}$ To simplify the given expression, $ \sqrt[3]{25(-3)^4(5)^3} ,$ use the laws of exponents to simplify the radicand. Then, find a factor of the radicand that is a perfect power of the index. Finally, extract the root of the factor that is a perfect power of the root. $\bf{\text{Solution Details:}}$ Since $25=5^2,$ the expression above is equivalent to \begin{array}{l}\require{cancel} \sqrt[3]{(5)^2(-3)^4(5)^3} .\end{array} Using the Product Rule of the laws of exponents which is given by $x^m\cdot x^n=x^{m+n},$ the expression above is equivalent to \begin{array}{l}\require{cancel} \sqrt[3]{(5)^{2+3}(-3)^4} \\\\= \sqrt[3]{(5)^{5}(-3)^4} .\end{array} Factoring the expression that is a perfect power of the index and then extracting the root result to \begin{array}{l}\require{cancel} \sqrt[3]{(5)^{3}(-3)^3\cdot\left[(5)^2(-3)^1 \right]} \\\\= \sqrt[3]{\left[5\cdot(-3)\right]^3\cdot\left[25(-3) \right]} \\\\= 5\cdot(-3)\sqrt[3]{25(-3)} \\\\= -15\sqrt[3]{-75} .\end{array}
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