College Algebra 7th Edition

Published by Brooks Cole
ISBN 10: 1305115546
ISBN 13: 978-1-30511-554-5

Chapter P, Prerequisites - Chapter P Review - Exercises - Page 77: 61



Work Step by Step

Since $x^6 = (x^2)^3$, the given expression is equivalent to: $=(x^2)^3-1$ The expression above is a difference of two cubes. Factor using the formula $a^3-b^3=(a-b)(a^2+ab+b^2)$ where $a=x^2$ and $b=1$ to obtain: $=(x^2-1)[(x^2)^2+x^2(1) + 1^2] \\=(x^2-1)(x^4+x^2+1)$ The first factor above is a difference of two squares. Factor using the formula $a^2-b^2=(a-b)(a+b)$ with $a=x$ and $b=1$ to obtain: $=(x-1)(x+1)(x^4+x^2+1)$
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