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: 45

Answer

$(x^4-yz^4)(x^8+x^4yz^4+y^2z^8)$

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, $ x^{12}-y^3z^{12} ,$ is \begin{array}{l} (x^4-yz^4)[ (x^4)^2-(x^4)(-yz^4)+(-yz^4)^2] \\\\= (x^4-yz^4)(x^8+x^4yz^4+y^2z^8) .\end{array}
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