Find the binomial expansion of #(3x-5/x^3)^7# in ascending power of #x#?

1 Answer
Feb 28, 2018

#-78125x^-21+328125x^-17+590625x^-13+590625x^-9-354375^-5+127575x^-1-25515x^3+2187x^7#

Explanation:

The binomial expansion for #(a+b)^7# is given by

#(a+b)^7 = a^7+7a^6b+21a^5b^2+35a^4b^3+35a^3b^4+21a^2b^5+7ab^6+b^7#

Choosing #a=3x# and #b=-5/x^3# gives

#(3x-5/x^3)^7 = (3x)^7+7(3x)^6(-5/x^3)+21(3x)^5(-5/x^3)^2+35(3x)^4(-5/x^3)^3+35(3x)^3(-5/x^3)^4+21(3x)^2(-5/x^3)^5+7(3x)(-5/x^3)^6+(-5/x^3)^7#
# = 2187x^7-25515x^{6-3}+127575x^{5-3times2} -354375x^{4-3times3}+590625x^{3-3times4}-590625x^{2-3times5}+328125x^{1-3times6}-78125x^{-3 times 7}#
#= -78125x^-21+328125x^-17+590625x^-13+590625x^-9-354375^-5+127575x^-1-25515x^3+2187x^7#