Intermediate Algebra (6th Edition)

Published by Pearson
ISBN 10: 0321785045
ISBN 13: 978-0-32178-504-6

Chapter 5 - Section 5.7 - Factoring by Special Products - Exercise Set - Page 309: 55

Answer

$\left(n-\dfrac{1}{3}\right)\left(n^2+\dfrac{n}{3}+\dfrac{1}{9}\right)$

Work Step by Step

Using $a^3+b^3=(a+b)(a^2-ab+b^2)$ or the factoring of the sum of 2 cubes, then, \begin{array}{l} n^3-\dfrac{1}{27} \\\\= \left[(n)+\left(-\dfrac{1}{3} \right) \right]\left[(n)^2-n\left(-\dfrac{1}{3} \right)+\left(-\dfrac{1}{3}\right) \right] \\\\= \left(n-\dfrac{1}{3}\right)\left(n^2+\dfrac{n}{3}+\dfrac{1}{9}\right) .\end{array}
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