Intermediate Algebra (6th Edition)

Published by Pearson
ISBN 10: 0321785045
ISBN 13: 978-0-32178-504-6

Chapter 5 - Section 5.7 - Factoring by Special Products - Exercise Set - Page 309: 33



Work Step by Step

Factoring the $GCF= y^2 $ results to $ y^2(27-x^3) $. Using $a^3+b^3=(a+b)(a^2-ab+b^2)$, or the factoring of the sum/difference of two cubes, then, \begin{array}{l} y^2(27-x^3) \\= y^2[3+(-x)][(3)^2-(3)(-x)+(-x)^2) \\= y^2(3-x)(9+3x+x^2) .\end{array}
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