Intermediate Algebra: Connecting Concepts through Application

Published by Brooks Cole
ISBN 10: 0-53449-636-9
ISBN 13: 978-0-53449-636-4

Chapter 3 - Exponents, Polynomials and Functions - 3.5 Special Factoring Techniques - 3.5 Exercises: 29

Answer

$(r+2)(r-2)(r^2-2r+4)(r^2+2r+4)$

Work Step by Step

$\bf{\text{Solution Outline:}}$ To factor the given expression, $ r^6-64 ,$ use the factoring of the difference of $2$ squares. Then use the factoring of the sum or difference of $2$ cubes. $\bf{\text{Solution Details:}}$ The expressions $ r^6 $ and $ 64 $ are both perfect squares (the square root is exact) and are separated by a minus sign. Hence, $ r^6-64 ,$ 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} (r^3)^2-(8)^2 \\\\= (r^3+8)(r^3-8) .\end{array} The expressions $ r^3 $ and $ 8 $ are both perfect cubes (the cube root is exact). Hence, $ r^3+8 $ 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} [(r)^3+(2)^3](r^3-8) \\\\= [(r+2)(r^2-r(2)+(2)^2](r^3-8) \\\\= (r+2)(r^2-2r+4)(r^3-8) .\end{array} The expressions $ r^3 $ and $ 8 $ are both perfect cubes (the cube root is exact). Hence, $ r^3-8 $ 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} (r+2)(r^2-2r+4)[(r)^3-(2)^3] \\\\= (r+2)(r^2-2r+4)(r-2)[(r)^2+r(2)+(2)^2] \\\\= (r+2)(r^2-2r+4)(r-2)(r^2+2r+4) \\\\= (r+2)(r-2)(r^2-2r+4)(r^2+2r+4) .\end{array}
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