## Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)

$8\left( t-1 \right)(t^2+t+1)$
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}