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: 11

Answer

$(x-4)(x+4x+16)$

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, $x^3-64 ,$ is \begin{array}{l} (x-4)[(x)^2-(x)(-4)+(-4)^2] \\\\= (x-4)[x+4x+16] \\\\= (x-4)(x+4x+16) .\end{array}

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