## Intermediate Algebra (6th Edition)

Published by Pearson

# Chapter 5 - Section 5.7 - Factoring by Special Products - Exercise Set - Page 309: 54

#### Answer

$x^2y^3(y^2+1)(y^4-y^2+1)$

#### Work Step by Step

Factoring the $GCF= x^2y^3$ results to $x^2y^3(y^6+1)$. 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} x^2y^3(y^6+1) \\= x^2y^3[(y^2)+(1)][(y^2)^2-(y^2)(1)+(1)^2] \\= x^2y^3(y^2+1)(y^4-y^2+1) .\end{array}

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