Introductory Algebra for College Students (7th Edition)

Published by Pearson
ISBN 10: 0-13417-805-X
ISBN 13: 978-0-13417-805-9

Chapter 6 - Section 6.4 - Factoring Special Forms - Exercise Set: 84

Answer

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

Work Step by Step

The given binomial can be written as: $(3y)^3-1^3$ The binomial above is a difference of two cubes. RECALL: A difference of two cubes can be factored using the formula: $a^3-b^3=(a-b)(a^2+ab+b^2)$ Use the formula above to factor the sum of two cubes 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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