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Assuming that, as suggested by tmonteil, you mean a function per cube root branch, the following one-liner :
sage: (f1,f2,f3)=tuple(((1/u)*1/(x+1)^3).function(x) for u in (x^3==1).roots(multiplicities=False))
does what you mean :
sage: [f1, f2, f3]
[x |--> 1/((x + 1)^3*(1/2*I*sqrt(3) - 1/2)),
x |--> 1/((x + 1)^3*(-1/2*I*sqrt(3) - 1/2)),
x |--> (x + 1)^(-3)]
But remember that the expression 1/(1+x)^(1/3) will evaluate to some cube root of x+1, with no specification of which one :
sage: ((x+1)^(1/3)).subs(x=7)
2
sage: ((x+1)^(1/3)).subs(x=-9)
2*(-1)^(1/3)
sage: ((x+1)^(1/3)).subs(x=-9).n()
1.00000000000000 + 1.73205080756888*I
Caveat emptor... HTH,
