Answer
$8\left( t-1 \right)(t^2+t+1)$
Work Step by Step
Factoring the $GCF=
8
,$ the given expression, $
8t^3-8
,$ is equivalent to
\begin{array}{l}
8(t^3-1)
.\end{array}
Using $a^3+b^3=(a+b)(a^2-ab+b^2)$ or the factoring of 2 cubes, the factored form of the expression, $
8(t^3-1)
,$ is
\begin{array}{l}
8\left( t-1 \right)\left[ (t)^2-(t)\left( -1 \right)+\left( -1 \right)^2 \right]
\\\\=
8\left( t-1 \right)(t^2+t+1)
.\end{array}
