Answer
$(x^2-y)(x^4+x^2y+y^2)$
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^6-y^3
\\=
[(x^2)+(-y)][(x^2)^2-(x^2)(-y)+(-y)^2]
\\=
(x^2-y)(x^4+x^2y+y^2)
.\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: 47
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