Intermediate Algebra (12th Edition)

Published by Pearson
ISBN 10: 0321969359
ISBN 13: 978-0-32196-935-4

Chapter 5 - Section 5.4 - A General Approach to Factoring - 5.4 Exercises: 29

Answer

$(p+1)(p^2-p+1)$

Work Step by Step

$\bf{\text{Solution Outline:}}$ To factor the given expression, $ p^3+1 ,$ use the factoring of the sum or difference of $2$ cubes. $\bf{\text{Solution Details:}}$ The expressions $ p^3 $ and $ 1 $ are both perfect cubes (the cube root is exact). Hence, $ p^3+1 $ is a $\text{ sum }$ of $2$ cubes. Using the factoring of the sum or difference of $2$ cubes which is given by $a^3+b^3=(a+b)(a^2-ab+b^2)$ or by $a^3-b^3=(a-b)(a^2+ab+b^2)$ the expression above is equivalent to \begin{array}{l}\require{cancel} (p)^3+(1)^3 \\\\= (p+1)[(p)^2-p(1)+(1)^2] \\\\= (p+1)(p^2-p+1) .\end{array}
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