Intermediate Algebra (12th Edition)

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

Chapter 5 - Chapter 5 Test - Page 362: 13


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

Work Step by Step

$\bf{\text{Solution Outline:}}$ To factor the given expression, $ 27x^6+1 ,$ use the factoring of the sum of $2$ cubes. $\bf{\text{Solution Details:}}$ The expressions $ 27x^6 $ and $ 1 $ are both perfect cubes (the cube root is exact). Hence, $ 27x^6+1 ,$ is a sum of $2$ cubes. Using the factoring of the sum of $2$ cubes which is given by $a^3+b^3=(a+b)(a^2-ab+b^2),$ the expression above is equivalent to \begin{array}{l}\require{cancel} (3x^2)^3+(1 )^3 \\\\= (3x^2+1 )[(3x^2)^2-3x^2(1 )+(1 )^2] \\\\= (3x^2+1 )(9x^4-3x^2+1) .\end{array}
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