Intermediate Algebra for College Students (7th Edition)

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

Chapter 6 - Section 6.6 - Rational Equations - Exercise Set - Page 463: 44


Solution set: $\{-5\}$

Work Step by Step

$x^{3}-1$ is a difference of cubes, $x^{3}-1^{3}=(x-1)(x^{2}+x+1)$ $x^{2}+x+1$ has no real zeros as $b^{2}-4ac=1-4$ ... is negative. LCD = $(x-1)(x^{2}+x+1)$ Exclude solutions which yield 0 in denominators: $x\not\in\{1\}\qquad (*)$ Multiply with the LCD, $(x-1)(x^{2}+x+1)=x^{3}-1$ $2+4=-1(x-1)$ $6=-x+1$ $5=-x$ $ x=-5\qquad$ ...satisfies (*). Solution set: $\{-5\}$
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