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: 61

Answer

$(3-10x^3)(9+30x^3+100x^6)$

Work Step by Step

$\bf{\text{Solution Outline:}}$ To factor the given expression, $ 27-1000x^9 ,$ use the factoring of the sum/difference of $2$ cubes. $\bf{\text{Solution Details:}}$ Using the factoring of the sum/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} (3-10x^3)[(3)^2+3(10x^3)+(10x^3)^2] \\\\= (3-10x^3)(9+30x^3+100x^6) .\end{array}
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.