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: 49

Answer

$32\sqrt[3]{2}$

Work Step by Step

$\bf{\text{Solution Outline:}}$ To simplify the given expression, $ \sqrt[3]{16(-2)^4(2)^8} ,$ use the laws of exponents to simplify the radicand. $\bf{\text{Solution Details:}}$ Since $(-2)^4=16$ and $2^8=256=16^2$ the expression above is equivalent to \begin{array}{l}\require{cancel} \sqrt[3]{16(16)(16)^2} .\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]{16^{1+1+2}} \\\\= \sqrt[3]{16^{4}} \\\\= \sqrt[3]{(2^4)^{4}} .\end{array} Using the Power Rule of the laws of exponents which is given by $\left( x^m \right)^p=x^{mp},$ the expression above is equivalent to \begin{array}{l}\require{cancel} \sqrt[3]{2^{4(4)}} \\\\= \sqrt[3]{2^{16}} \\\\= \sqrt[3]{2^{15}\cdot2^1} \\\\= \sqrt[3]{(2^{5})^3\cdot2} \\\\= 2^{5}\sqrt[3]{2} \\\\= 32\sqrt[3]{2} .\end{array}
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