Intermediate Algebra for College Students (7th Edition)

Published by Pearson
ISBN 10: 0-13417-894-7
ISBN 13: 978-0-13417-894-3

Chapter 5 - Section 5.5 - Factoring Special Forms - Exercise Set: 80

Answer

$(3y+1)(9y^2-3y+1)$

Work Step by Step

The given binomial can be written as: $(3y)^3+1^3$ This binomial is a sum of two cubes. RECALL: $a^3+b^3=(a+b)(a^2-ab+b^2)$ Factor the difference of two cubes using the formula above with $a=3y$ and $b=1$ to obtain: $=(3y+1)[(3y)^2-3y(1)+1^2) \\=(3y+1)(9y^2-3y+1)$
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