Answer
$(x-2)(x^2+2x+4)$
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^3-8
\\=
(x-2)[(x)^2+(x)(2)+(-2)^2)
\\=
(x-2)(x^2+2x+4)
.\end{array}
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