Answer
True
Work Step by Step
$x-1$ is a factor of $x^3 -1$, so it does have a remainder of 0.
Elementary Algebra
by Kaufmann, Jerome; Schwitters, Karen;
Published by
Cengage Learning
ISBN 10:
1285194055
ISBN 13:
978-1-28519-405-9
Chapter 5 - Exponents and Polynomials - 5.5 - Dividing by Binomials - Concept Quiz 5.5 - Page 217: 7
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