Intermediate Algebra (12th Edition)

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

Chapter 5 - Review Exercises: 31

Answer

$(5x-1)(25x^2+5x+1)$

Work Step by Step

$\bf{\text{Solution Outline:}}$ To factor the given expression, $ 125x^3-1 ,$ use the factoring of the sum/difference of $2$ cubes. $\bf{\text{Solution Details:}}$ The expressions $ 125x^3 $ and $ 1 $ are both perfect cubes (the cube root is exact). Hence, $ 125x^3-1 $ 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} (5x)^3-(1)^3 \\\\= (5x-1)[(5x)^2+5x(1)+(1)^2] \\\\= (5x-1)(25x^2+5x+1) .\end{array}
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