#### Answer

$(x+1)(x-1)(x-2)(x^2+2x+4)$

#### Work Step by Step

$\bf{\text{Solution Outline:}}$
To factor the given expression, $
x^5-x^3-8x^2+8
,$ use factoring by grouping. This results to a factor that is a difference of $2$ squares and a factor that is a difference of $2$ cubes. Use the appropriate factoring method for each factor.
$\bf{\text{Solution Details:}}$
Using factoring by grouping, the given expression is equivalent to
\begin{array}{l}\require{cancel}
(x^5-x^3)-(8x^2-8)
\\\\=
x^3(x^2-1)-8(x^2-1)
\\\\=
(x^2-1)(x^3-8)
.\end{array}
The expressions $
x^2
$ and $
1
$ are both perfect squares (the square root is exact) and are separated by a minus sign. Hence, $
x^2-1
,$ is a difference of $2$ squares. Using the factoring of the difference of $2$ squares which is given by $a^2-b^2=(a+b)(a-b),$ the expression above is equivalent to
\begin{array}{l}\require{cancel}
[(x)^2-(1)^2](x^3-8)
\\\\=
(x+1)(x-1)(x^3-8)
.\end{array}
The expressions $
x^3
$ and $
8
$ are both perfect cubes (the cube root is exact). Hence, $
x^3-8
$ is a $\text{
difference
}$ of $2$ cubes. Using the factoring of the sum or difference of $2$ cubes which is given by $a^3+b^3=(a+b)(a^2-ab+b^2)$ or by $a^3-b^3=(a-b)(a^2+ab+b^2)$ the expression above is equivalent to
\begin{array}{l}\require{cancel}
(x+1)(x-1)[(x)^3-(2)^3]
\\\\=
(x+1)(x-1)[(x-2)(x^2+2x+4)]
\\\\=
(x+1)(x-1)(x-2)(x^2+2x+4)
.\end{array}