Intermediate Algebra (12th Edition)

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

Chapter 5 - Review Exercises: 35

Answer

$(2b)(3a^2+b^2)$

Work Step by Step

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