Intermediate Algebra (12th Edition)

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

Chapter 5 - Review Exercises: 32

Answer

$(m+1)(m-1)(m^2-m+1)(m^2+m+1)$

Work Step by Step

$\bf{\text{Solution Outline:}}$ To factor the given expression, $ m^6-1 ,$ use the factoring of the difference of $2$ squares. Then use the factoring of the sum/difference of $2$ cubes to factor further. $\bf{\text{Solution Details:}}$ The expressions $ m^6 $ and $ 1 $ are both perfect squares (the square root is exact) and are separated by a minus sign. Hence, $ m^6-1 ,$ is a difference of $2$ squares. Using the factoring of the difference of $2$ squares which is given by $a^2-b^2=(a+b)(a-b),$ the expression above is equivalent to \begin{array}{l}\require{cancel} (m^3)^2-(1)^2 \\\\= (m^3+1)(m^3-1) .\end{array} The expressions $ m^3 $ and $ 1 $ are both perfect cubes (the cube root is exact). Hence, $ m^3+1 $ is a $\text{ sum }$ 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} [(m)^3+(1)^3](m^3-1) \\\\= [(m+1)(m^2-m+1)](m^3-1) \\\\= (m+1)(m^2-m+1)(m^3-1) .\end{array} The expressions $ m^3 $ and $ 1 $ are both perfect cubes (the cube root is exact). Hence, $ m^311 $ 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} (m+1)(m^2-m+1)[(m)^3-(1)^3] \\\\= (m+1)(m^2-m+1)[(m-1)(m^2+m+1)] \\\\= (m+1)(m^2-m+1)(m-1)(m^2+m+1) \\\\= (m+1)(m-1)(m^2-m+1)(m^2+m+1) .\end{array}
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