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

Published by Pearson

# Chapter 5 - Polynomials and Factoring - 5.5 Factoring Sums or Differences of Cubes - 5.5 Exercise Set - Page 339: 63

#### Answer

$(t-8)(t-1)(t^2+t+1)$

#### Work Step by Step

Grouping the first and second terms and the third and fourth terms, the given expression is equivalent to \begin{array}{l}\require{cancel} t^4-8t^3-t+8 \\\\= (t^4-8t^3)-(t-8) .\end{array} Factoring the $GCF$ in each group results to \begin{array}{l}\require{cancel} t^3(t-8)-(t-8) .\end{array} Factoring the $GCF= (t-8)$ of the entire expression above results to \begin{array}{l}\require{cancel} (t-8)(t^3-1) .\end{array} The expressions $t^3$ and $1$ are both perfect cubes (the cube root is exact). Hence, $t^3-1$ is a $\text{ difference }$ of $2$ cubes. Using the factoring of the sum or difference of $2$ cubes which is given by $a^3+b^3=(a+b)(a^2-ab+b^2)$ or by $a^3-b^3=(a-b)(a^2+ab+b^2)$ the expression above is equivalent to \begin{array}{l}\require{cancel} (t-8)[(t)^3-(1)^3] \\\\= (t-8)(t-1)[(t)^2+t(1)+(1)^2] \\\\= (t-8)(t-1)(t^2+t+1) .\end{array}

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