College Algebra 7th Edition

Published by Brooks Cole
ISBN 10: 1305115546
ISBN 13: 978-1-30511-554-5

Chapter P, Prerequisites - Section P.6 - Factoring - P.6 Exercises - Page 44: 120

Answer

See the explanation

Work Step by Step

a. $A^2-1=(A-1)(A+1)$ $=A^2+A-A-1$ $=A^2-1$ $A^3-1=(A-1)(A^2+A+1)$ $=A^3+A^2+A-A^2-A-1$, $=A^3-1$ $A^4=(A-1)(A^3+A^2+A+1)$ $=A^4+A^3+A^2+A-A^3-A^2-A-1$ $=A^4-1$ b.$A^5-1=(A-1)(A^4+A^3+A^2+A+1)$ $=A^5+A^4+A^3+A^2+A-A^4-A^3-A^2-A-1$ $=A^5-1$ Therefore, $A^n-1=(A-1)(A^{n-1}+A^{n-2}+A^{n-3}+A^{n-4}+....+A+1)$
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