Intermediate Algebra (12th Edition)

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

Chapter 5 - Section 5.3 - Special Factoring - 5.3 Exercises - Page 343: 53



Work Step by Step

$\bf{\text{Solution Outline:}}$ To factor the given expression, $ 343p^3+125q^3 ,$ use the factoring of the sum/difference of $2$ cubes. $\bf{\text{Solution Details:}}$ Using $a^3+b^3=(a+b)(a^2-ab+b^2)$ or $a^3-b^3=(a-b)(a^2+ab+b^2)$ or the factoring of the sum/difference of $2$ cubes, the expression above is equivalent to \begin{array}{l}\require{cancel} (7p+5q)[(7p)^2-7p(5q)+(5q)^2] \\\\= (7p+5q)(49p^2-35pq+25q^2) .\end{array}
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