Answer
$8t\left( t-1 \right)\left( {{t}^{2}}+t+1 \right)$
Work Step by Step
$8{{t}^{4}}-8t$
Factor our 8t:
$8t\left( {{t}^{3}}-1 \right)$
Rewrite the equation as a difference of cubes:
$8t\left( {{t}^{3}}-1 \right)=8t\left( {{t}^{3}}-{{1}^{3}} \right)$
Apply the difference of cubes equation as shown below,
$\begin{align}
& 8t\left( {{t}^{3}}-1 \right)=8t\left( {{t}^{3}}-{{1}^{3}} \right) \\
& =8t\left( t-1 \right)\left( {{t}^{2}}+t+1 \right)
\end{align}$
Check:
$\begin{align}
& 8t\left( t-1 \right)\left( {{t}^{2}}+t+1 \right)=8t\left( t\left( {{t}^{2}}+t+1 \right)-1\left( {{t}^{2}}+t+1 \right) \right) \\
& =8t\left( {{t}^{3}}+{{t}^{2}}+t-{{t}^{2}}-t-1 \right) \\
& =8t\left( {{t}^{3}}-1 \right) \\
& =8{{t}^{4}}-8t
\end{align}$
Thus, the answer checks.
