#### Answer

$(t-8)(t-1)(t^2+t+1)$

#### Work Step by Step

Grouping the first and second terms and the third and fourth terms, the given expression is equivalent to
\begin{array}{l}\require{cancel}
t^4-8t^3-t+8
\\\\=
(t^4-8t^3)-(t-8)
.\end{array}
Factoring the $GCF$ in each group results to
\begin{array}{l}\require{cancel}
t^3(t-8)-(t-8)
.\end{array}
Factoring the $GCF=
(t-8)
$ of the entire expression above results to
\begin{array}{l}\require{cancel}
(t-8)(t^3-1)
.\end{array}
The expressions $
t^3
$ and $
1
$ are both perfect cubes (the cube root is exact). Hence, $
t^3-1
$ 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}
(t-8)[(t)^3-(1)^3]
\\\\=
(t-8)(t-1)[(t)^2+t(1)+(1)^2]
\\\\=
(t-8)(t-1)(t^2+t+1)
.\end{array}