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

Published by Pearson

# Chapter 5 - Polynomials and Factoring - 5.5 Factoring Sums or Differences of Cubes - 5.5 Exercise Set - Page 338: 20

#### Answer

$(3-2t)(9+6t+4t^2)$

#### 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, $27-8t^3 ,$ is \begin{array}{l} (3-2t)[(3)^2-(3)(-2t)+(-2t)^2] \\\\= (3-2t)(9+6t+4t^2) .\end{array}

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