Intermediate Algebra (6th Edition)

Published by Pearson
ISBN 10: 0321785045
ISBN 13: 978-0-32178-504-6

Chapter 6 - Section 6.4 - Dividing Polynomials: Long Division and Synthetic Division - Exercise Set - Page 371: 63

Answer

$x^3-y^3=(x-1)(x^2+x+1)$

Work Step by Step

The given expression can be written as: $=x^3-1^3$ RECALL: A sum or difference of two cubes can be factored using the following: (i) $a^3-b^3=(a-b)(a^2+ab+b^2)$ (ii) $a^3+b^3 = (a+b)(a^2-ab+b^2)$ Use formula (1) above with $a=x$ and $b=1$ to have: $=(x-1)(x^2+x(1)+1^2) \\=(x-1)(x^2+x+1)$
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