# Chapter 5 - Section 5.3 - Special Factoring - 5.3 Exercises - Page 343: 39

$(6-t)(36+6t+t^2)$

#### Work Step by Step

$\bf{\text{Solution Outline:}}$ To factor the given expression, $216-t^3 ,$ use the factoring of the sum/difference of $2$ cubes. $\bf{\text{Solution Details:}}$ Using $(a\pm b)(a^2\mp ab+b^2)$ or the factoring of the sum/difference of $2$ cubes, the expression above is equivalent to \begin{array}{l}\require{cancel} (6-t)[(6)^2+6(t)+(t)^2] \\\\= (6-t)(36+6t+t^2) .\end{array}

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