# Chapter 9 - Inequalities and Problem Solving - 9.2 Intersections, Unions, and Compound Inequalities - 9.2 Exercise Set - Page 591: 97

$10(c+1)(c^2-c+1)(c-1)(c^2+c+1)$

#### Work Step by Step

$\bf{\text{Solution Outline:}}$ Get the $GCF$ of the given expression, $10c^6-10 .$ Then use the factoring of the difference of $2$ squares and the factoring of the sum and difference of $2$ cubes. $\bf{\text{Solution Details:}}$ The $GCF$ of the terms is $10$ since it is the highest number that can evenly divide (no remainder) all the given terms. Factoring the $GCF,$ the expression above is equivalent to \begin{array}{l}\require{cancel} 10c^6-10 \\\\= 10(c^6-1) .\end{array} The expressions $c^6$ and $1$ are both perfect squares (the square root is exact) and are separated by a minus sign. Hence, $c^6-1$ is a difference of $2$ squares. Using the factoring of the difference of $2$ squares which is given by $a^2-b^2=(a+b)(a-b),$ the expression above is equivalent to \begin{array}{l}\require{cancel} 10(c^6-1) \\\\= 10(c^3+1)(c^3-1) .\end{array} The expressions $c^3$ and $1$ are both perfect cubes (the cube root is exact). Hence, $c^3+1$ is a $\text{ sum }$ 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} 10(c^3+1)(c^3-1) \\\\= 10[(c)^3+(1)^3](c^3-1) \\\\= 10(c+1)[(c)^2-c(1)+(1)^2](c^3-1) \\\\= 10(c+1)(c^2-c+1)(c^3-1) .\end{array} The expressions $c^3$ and $1$ are both perfect cubes (the cube root is exact). Hence, $c^3-1$ 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} 10(c+1)(c^2-c+1)(c^3-1) \\\\= 10(c+1)(c^2-c+1)[(c)^3-(1)^3] \\\\= 10(c+1)(c^2-c+1)(c-1)[(c)^2+c(1)+(1)^2] \\\\= 10(c+1)(c^2-c+1)(c-1)(c^2+c+1) .\end{array}

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