Intermediate Algebra for College Students (7th Edition)

Published by Pearson
ISBN 10: 0-13417-894-7
ISBN 13: 978-0-13417-894-3

Chapter 8 - Section 8.4 - Equations Quadratic in Form - Exercise Set - Page 635: 25



Work Step by Step

Since, $u^2-u-6=0$ Factorize the expression as follows: $(u+2)(u-3)=0$ or, $u=${$-2,3$} Replace $u$ with $x^{1/3}$, we have $x^{1/3}=-2\implies x=-2^3$ $x=-8$ and $x^{1/3}=3\implies x=-3^3$ $x=27$ Hence, our solution set is $x=${$-8,27$}
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