Precalculus (6th Edition)

Published by Pearson
ISBN 10: 013421742X
ISBN 13: 978-0-13421-742-0

Chapter R - Review of Basic Concepts - Test: 20

Answer

$\color{blue}{(1-3x^2)(1+3x^2+9x^4)}$

Work Step by Step

With $27x^6=(3x^2)^3$ and $1=1^3$, the given expression is equivalent to: $=1^3-(3x^2)^3$ Factor using the formula $a^3-b^3=(a-b)(a^2+ab+b^2)$ with $a=1$ and $b=3x^2$ to obtain: $=(1-3x^2)[1^2+1(3x^2)+(3x)^2] \\=\color{blue}{(1-3x^2)(1+3x^2+9x^4)}$
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