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 - Page 372: 79

Answer

$(2y+1)(4y^2-2y+1)$

Work Step by Step

The formula for factoring the sum of two cubes is: $A^3+B^3=(A+B)(A^2-AB+B^2)$. The formula for factoring the difference of two cubes is: $A^3-B^3=(A-B)(A^2+AB+B^2)$. Hence here: $8y^3+1=\\=(2y)^3+1^3\\=(2y+1)((2y)^2-2y\cdot1+1^2)\\=(2y+1)(4y^2-2y+1)$
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