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: 80

Answer

$\color{blue}{b(b^2+9b+27)}$

Work Step by Step

With $27=3^3$, the given polynomial is equivalent to: $=(b+3)^3-3^3$ Factor using the formula $a^3-b^3=(a-b)(a^2+ab+b^2)$ with $a=b+3$ and $b=3$ to obtain: $=(b+3-3)[(b+3)^2+(b+3)(3) + 3^2] \\=(b+0)[(b^2+6b+9)+3b+9+9] \\=b(b^2+6b+9+3b+18) \\=\color{blue}{b(b^2+9b+27)}$
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