Answer
$(x-y^2)(x^2+xy^2+y^4)$
Work Step by Step
Using $a^3+b^3=(a+b)(a^2-ab+b^2)$ or the factoring of the difference of 2 cubes, then,
\begin{array}{l}
x^3-y^6
\\=
[(x)+(-y^2)][(x)^2-(x)(-y^2)+(-y^2)^2]
\\=
(x-y^2)(x^2+xy^2+y^4)
.\end{array}
Intermediate Algebra (6th Edition)
by Martin-Gay, Elayn
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: 48
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