Intermediate Algebra (12th Edition)

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

Chapter 5 - Review Exercises: 30

Answer

$(r+3)(r^2-3r+9)$

Work Step by Step

$\bf{\text{Solution Outline:}}$ To factor the given expression, $ r^3+27 ,$ use the factoring of the sum/difference of $2$ cubes. $\bf{\text{Solution Details:}}$ The expressions $ r^3 $ and $ 27 $ are both perfect cubes (the cube root is exact). Hence, $ r^3+27 $ 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} (r)^3+(3)^3 \\\\= (r+3)[(r)^2-r(3)+(3)^2] \\\\= (r+3)(r^2-3r+9) .\end{array}
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