## Intermediate Algebra (12th Edition)

Published by Pearson

# Chapter 5 - Section 5.3 - Special Factoring - 5.3 Exercises - Page 343: 46

#### Answer

$(3y+1)(9y^2-3y+1)$

#### Work Step by Step

$\bf{\text{Solution Outline:}}$ To factor the given expression, $27y^3+1 ,$ use the factoring of the sum/difference of $2$ cubes. $\bf{\text{Solution Details:}}$ Using $(a\pm b)(a^2\mp ab+b^2)$ or the factoring of the sum/difference of $2$ cubes, the expression above is equivalent to \begin{array}{l}\require{cancel} (3y+1)[(3y)^2-3y(1)+(1)^2] \\\\= (3y+1)(9y^2-3y+1) .\end{array}

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