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 5.6 - A General Factoring Strategy - Exercise Set - Page 379: 59

Answer

$x^3y^3(xy-1)(x^2y^2+xy+1)$

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: $x^6y^6-x^3y^3=\\=x^3y^3(x^3y^3-1)\\=x^3y^3((xy)^3-1^3)\\=x^3y^3(xy-1)((xy)^2+xy+1^2)\\=x^3y^3(xy-1)(x^2y^2+xy+1)$
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