# Chapter 5 - Polynomials and Factoring - 5.5 Factoring Sums or Differences of Cubes - 5.5 Exercise Set - Page 338: 25

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

#### 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}

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