Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)

Published by Pearson
ISBN 10: 0-32184-874-8
ISBN 13: 978-0-32184-874-1

Chapter 4 - Polynomials - 4.6 Special Products - 4.6 Exercise Set - Page 277: 83

Answer

$a^3+1$

Work Step by Step

Using $a(b+c)=ab+ac$ or the Distributive Property, the given expression, $ (a+1)(a^2-a+1) $, is equivalent to \begin{array}{l}\require{cancel} a(a^2)+a(-a)+a(1)+1(a^2)+1(-a)+1(1) \\\\= a^3-a^2+a+a^2-a+1 \\\\= a^3+(-a^2+a^2)+(a-a)+1 \\\\= a^3+0a^2+0a+1 \\\\= a^3+1 .\end{array}
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