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

Published by Pearson

# Chapter R - Elementary Algebra Review - R.5 Polynomials and Factoring - R.5 Exercise Set - Page 970: 26

#### Answer

$\left(\dfrac{1}{3}+x\right)\left(\dfrac{1}{9}-\dfrac{1}{3}x+x^2\right)$

#### Work Step by Step

The expressions $\dfrac{1}{27}$ and $x^3$ are both perfect cubes (the cube root is exact). Hence, $\dfrac{1}{27}+x^3$ 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} \left(\dfrac{1}{3}\right)^3+(x)^3 \\\\= \left(\dfrac{1}{3}+x\right)\left[\left(\dfrac{1}{3}\right)^2-\dfrac{1}{3}(x)+(x)^2\right] \\\\= \left(\dfrac{1}{3}+x\right)\left(\dfrac{1}{9}-\dfrac{1}{3}x+x^2\right) .\end{array}

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