Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)

Published by Pearson
ISBN 10: 0-32184-874-8
ISBN 13: 978-0-32184-874-1

Chapter 5 - Polynomials and Factoring - 5.5 Factoring Sums or Differences of Cubes - 5.5 Exercise Set: 46

Answer

$(a^3-b^4c^5)(a^6+a^3b^4c^5+b^8c^{10})$

Work Step by Step

Using $a^3+b^3=(a+b)(a^2-ab+b^2)$ or the factoring of 2 cubes, the factored form of the given expression, $ a^{9}+b^{12}c^{15} ,$ is \begin{array}{l} (a^3-b^4c^5)[ (a^3)^2-(a^3)(-b^4c^5)+(-b^4c^5)^2] \\\\= (a^3-b^4c^5)(a^6+a^3b^4c^5+b^8c^{10}) .\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.