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: 56



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} p^3+\dfrac{1}{25} \\\\= \left[(p)+\left(\dfrac{1}{5} \right) \right]\left[(p)^2-p\left(\dfrac{1}{5} \right)+\left(\dfrac{1}{5}\right)^2 \right] \\\\= \left(p+\dfrac{1}{5}\right)\left(p^2-\dfrac{p}{5}+\dfrac{1}{25}\right) .\end{array}
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