Answer
$x^5-15x^4y+90x^3y^2-270x^2y^3+405xy^4-243y^5$
Work Step by Step
Use the formula for binomial expansion, we have:
$(x-3y)^5=(x)^5+5(x)^4(-3y)+10(x)^3(-3y)^2+10(x)^2(-3y)^3+5(x)(-3y)^4+(-3y)^5
=x^5-15x^4y+90x^3y^2-270x^2y^3+405xy^4-243y^5$
Precalculus (6th Edition)
by Lial, Margaret L.; Hornsby, John; Schneider, David I.; Daniels, Callie
Published by
Pearson
ISBN 10:
013421742X
ISBN 13:
978-0-13421-742-0
Chapter 11 - Further Topics in Algebra - Quiz - Page 1053: 7
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