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

Published by Pearson
ISBN 10: 0-32184-874-8
ISBN 13: 978-0-32184-874-1

Chapter R - Elementary Algebra Review - R.5 Polynomials and Factoring - R.5 Exercise Set: 8

Answer

$(y-1)(y^2+y+1)$

Work Step by Step

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