College Algebra (11th Edition)

Published by Pearson
ISBN 10: 0321671791
ISBN 13: 978-0-32167-179-0

Chapter R - Test: 20

Answer

$(1-3x^2)(1+3x^2+9x^4)$

Work Step by Step

$\bf{\text{Solution Outline:}}$ To factor the given expression, $ 1-27x^6 ,$ use the factoring of the sum or difference of $2$ cubes. $\bf{\text{Solution Details:}}$ The expressions $ 1 $ and $ 27x^6 $ are both perfect cubes (the cube root is exact). Hence, $ 1-27x^6 $ 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} 1^3-(3x^2)^3 \\\\= (1-3x^2)(1+3x^2+9x^4) .\end{array}
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