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: 12

Answer

$(t-3)(t^2+3t+9)$

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, $ t^3-27 ,$ is \begin{array}{l} (t-3)[(t)^2-(t)(-3)+(-3)^2] \\\\= (t-3)[t^2+3t+9] \\\\= (t-3)(t^2+3t+9) .\end{array}
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