Intermediate Algebra (12th Edition)

Published by Pearson
ISBN 10: 0321969359
ISBN 13: 978-0-32196-935-4

Chapter 5 - Section 5.4 - A General Approach to Factoring - 5.4 Exercises: 66

Answer

$(x-27)(x^2+3y^2+27x-81y-3xy+729)$

Work Step by Step

$\bf{\text{Solution Outline:}}$ To factor the given expression, $ (x-y)^3-(27-y)^3 ,$ use the factoring of the sum or difference of $2$ cubes. $\bf{\text{Solution Details:}}$ The expressions $ (x-y)^3 $ and $ (27-y)^3 $ are both perfect cubes (the cube root is exact). Hence, $ (x-y)^3-(27-y)^3 $ is a $\text{ difference }$ 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} [(x-y)-(27-y)][(x-y)^2+(x-y)(27-y)+(27-y)^2] \\\\= [x-y-27+y][(x-y)^2+(x-y)(27-y)+(27-y)^2] \\\\= [x-27][(x-y)^2+(x-y)(27-y)+(27-y)^2] .\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} [x-27][(x^2-2xy+y^2)+(x-y)(27-y)+(729-54y+y^2)] .\end{array} Using the FOIL Method which is given by $(a+b)(c+d)=ac+ad+bc+bd,$ the expression above is equivalent to\begin{array}{l}\require{cancel} [x-27][(x^2-2xy+y^2)+(27x-xy-27y+y^2)+(729-54y+y^2)] .\end{array} By combining like terms, the expression above is equivalent to \begin{array}{l}\require{cancel} [x-27][x^2+(y^2+y^2+y^2)+27x+(-27y-54y)+(-2xy-xy)+729] \\\\= (x-27)(x^2+3y^2+27x-81y-3xy+729) .\end{array}
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