How do you expand #(x^3-1/x^2)^3#?

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Answer 1 · 2018-06-12T12:19:40

#x^9-3x^4+3/x-1/x^6#

now

#(a+b)^3=a^3+3a^2b+3ab^2+b^3--(1)#

#(x^3-1/x^2)^3#

#a=x^3, b=-1/x^2#

substituting into #(1)#

#=(x^3)^3+3(x^3)^2(-1/x^2)+3x^3(-1/x^2)^2+(-1/x^2)^3#

now simplify

#=x^9-3x^6/x^2+3x^3/x^4-1/x^6#

#=x^9-3x^4+3/x-1/x^6#