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: 97



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: $8x^4-\frac{x}{8}=\\=\frac{1}{8}(64x^3-1)\\=\frac{1}{8}((4x)^3-1^3)\\=\frac{1}{8}(4x-1)((4x)^2+1(4x)+1^2)\\=\frac{1}{8}(4x-1)(16x^2+4x+1)$
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