## Intermediate Algebra (12th Edition)

Published by Pearson

# Chapter 5 - Chapter 5 Test - Page 362: 11

#### Answer

$( y-6 )( y^2+6y+36)$

#### Work Step by Step

$\bf{\text{Solution Outline:}}$ To factor the given expression, $y^3-216 ,$ use the factoring of the difference of $2$ cubes. $\bf{\text{Solution Details:}}$ The expressions $y^3$ and $216$ are both perfect cubes (the cube root is exact). Hence, $y^3-216 ,$ is a difference of $2$ cubes. Using the factoring of the sum or difference of $2$ cubes which is given by $a^3+b^3=(a+b)(a^2-ab+b^2)$ or by $a^3-b^3=(a-b)(a^2+ab+b^2)$ the expression above is equivalent to \begin{array}{l}\require{cancel} ( y)^3-(6 )^3 \\\\= ( y-6 )[( y)^2+ y(6 )+(6 )^2] \\\\= ( y-6 )( y^2+6y+36) .\end{array}

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