College Algebra 7th Edition

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

Chapter 8, Sequences and Series - Section 8.6 - The Binomial Theorem - 8.6 Exercises: 11

Answer

$(x^{2}y-1)^{5}=x^{10}y^{5}-5x^{8}y^{4}+10x^{6}y^{3}-10x^{4}y^{2}+5x^{2}y-1$

Work Step by Step

We use the values from the 5th row of Pascal's Triangle to find the coefficients: $(x^{2}y-1)^{5}=(x^{2}y)^{5}-5(x^{2}y)^{4}+10(x^{2}y)^{3}-10(x^{2}y)^{2}+5(x^{2}y)^1-1=x^{10}y^{5}-5x^{8}y^{4}+10x^{6}y^{3}-10x^{4}y^{2}+5x^{2}y-1$
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