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: 30

Answer

$(x+3)(x-3)(x^2-3x+9)(x^2+3x+9)$

Work Step by Step

$\bf{\text{Solution Outline:}}$ To factor the given expression, $ x^6-729 ,$ 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 $ x^6 $ and $ 729 $ are both perfect squares (the square root is exact) and are separated by a minus sign. Hence, $ x^6-729 ,$ 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} (x^3)^2-(27)^2 \\\\= (x^3+27)(x^3-27) .\end{array} The expressions $ x^3 $ and $ 27 $ are both perfect cubes (the cube root is exact). Hence, $ x^3+27 $ 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} [(x)^3+(3)^3](x^3-27) \\\\= (x+3)[(x)^2-x(3)+(3)^2](x^3-27) \\\\= (x+3)(x^2-3x+9)(x^3-27) .\end{array} The expressions $ x^3 $ and $ 27 $ are both perfect cubes (the cube root is exact). Hence, $ x^3-27 $ 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+3)(x^2-3x+9)[(x)^3-(3)^3] \\\\= (x+3)(x^2-3x+9)(x-3)[(x)^2+x(3)+(3)^2] \\\\= (x+3)(x^2-3x+9)(x-3)(x^2+3x+9) \\\\= (x+3)(x-3)(x^2-3x+9)(x^2+3x+9) .\end{array}
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