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 - Page 44: 82



Work Step by Step

With $125=5^3$, the given polynomial is equivalent to: $=5^3-(4a-b)^3$ Factor using the formula $a^3-b^3=(a-b)(a^2+ab+b^2)$ with $a=5$ and $b=4a-b$ to obtain: $=[5-(4a-b)][5^2+5(4a-b) + (4a-b)^2] \\=(5-4a+b)[25+20a-5b+(16a^2-8ab+b^2)] \\=(5-4a+b)(25+20a-5b+16a^2-8ab+b^2) \\=\color{blue}{(5-4a+b)(16a^2+b^2-8ab+20a-5b+25)}$
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