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

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

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

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