## College Algebra (11th Edition)

Published by Pearson

# Chapter R - Section R.3 - Polynomials - R.3 Exercises: 74

#### Answer

$z^3-9z^2+27z-27$

#### Work Step by Step

Using $(a\pm b)^3=a^3\pm3a^2b+3ab^2\pm b^3$ or the cube of a binomial, the expression, $(z-3)^3 ,$ is equivalent to \begin{array}{l}\require{cancel} (z)^3-3(z)^2(3)+3(z)(3)^2-(3)^3 \\\\= z^3-3(z^2)(3)+3(z)(9)-27 \\\\= z^3-9z^2+27z-27 .\end{array}

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