Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 10 - Section 10.5 - The Binomial Theorum - Exercise Set - Page 1092: 52



Work Step by Step

Using the Binomial Theorem, we can expand the expression as $(x^{2/3}-\frac{1}{\sqrt[3] x})^3=(x^{2/3}-x^{-1/3})^3=\binom30 (x^{2/3})^3(-x^{-1/3})^0+\binom31 (x^{2/3})^2(-x^{-1/3})^1+\binom32 (x^{2/3})^1(-x^{-1/3})^2+\binom33 (x^{2/3})^0(-x^{-1/3})^3 =x^2-3x+3-x^{-1}=x^2-3x-\frac{1}{x}+3$
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