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: 82

Answer

$(3x-2)(9x^2+6x+4)$

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: $27x^3-8=\\=(7x)^3-2^3\\=(3x-2)((3x)^2+3x\cdot2+2^2)\\=(3x-2)(9x^2+6x+4)$
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