## Intermediate Algebra (12th Edition)

Published by Pearson

# Chapter 5 - Section 5.3 - Special Factoring - 5.3 Exercises: 57

#### Answer

$(y+z+4)(y^2+2yz+z^2-4y-4z+16)$

#### Work Step by Step

$\bf{\text{Solution Outline:}}$ To factor the given expression, $(y+z)^3+64 ,$ use the factoring of the sum/difference of $2$ cubes. $\bf{\text{Solution Details:}}$ Using the factoring of the sum/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+z)+4][(y+z)^2-(y+z)(4)+(4)^2] \\\\= (y+z+4)[(y+z)^2-4y-4z+16] .\end{array} Using the square of a binomial which is given by $(a+b)^2=a^2+2ab+b^2$ or by $(a-b)^2=a^2-2ab+b^2,$ the expression above is equivalent to\begin{array}{l}\require{cancel} (y+z+4)[(y^2+2yz+z^2)-4y-4z+16] \\\\= (y+z+4)(y^2+2yz+z^2-4y-4z+16) .\end{array}

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