Answer
$(x+3)(x^2-3x+9)$
Work Step by Step
Using $a^3\pm b^3=(a\pm b)(a^2\mp ab+b^2)$, or the factoring of the sum/difference of two cubes, then,
\begin{array}{l}
x^3+27
\\=
(x+3)[(x)^2-(x)(3)+(3)^2)
\\=
(x+3)(x^2-3x+9)
.\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: 27
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