Precalculus (6th Edition)

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

Chapter R - Review of Basic Concepts - R.4 Factoring Polynomials - R.4 Exercises: 76

Answer

$\color{blue}{(2m-3n)(4m^2+6mn+9n^2)}$

Work Step by Step

With $8m^3=(2m)^3$ and $27n^3=(3n)^3$, the given polynomial is equivalent to: $=(2m)^3-(3n)^3$ Factor using the formula $a^3-b^3=(a-b)(a^2+ab+b^2)$ with $a=2m$ and $b=3n$ to obtain: $=(2m-3n)[(2m)^2+(2m)(3n) + (3n)^2] \\=\color{blue}{(2m-3n)(4m^2+6mn+9n^2)}$
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