Find the Laurent series of the function (z^3-z)^-1 in the region 1<|z-1|<2

Find the Laurent series of the function (z^3-z)^-1 in the region 1<|z-1|<2
Thanks for your help

Note that (z^3 - z)^(-1)
= 1/(z(z^2 - 1))
= (1/2) [1/(z - 1) + 1/(z + 1) - 2/z], via partial fractions
= (1/2) [1/(z - 1) + 1/((z - 1) + 2) - 2/((z - 1) + 1)]
= (1/2) [1/(z - 1) + (1/2) * 1/(1 + (z - 1)/2) - (2/(z - 1)) * (1 + 1/(z - 1))]
= (1/4) [2/(z - 1) + 1/(1 - -(z - 1)/2) - (4/(z - 1)) * 1/(1 - -1/(z - 1))]
= (1/4) [2/(z - 1) + Σ(n = 0 to ∞) (-(z - 1)/2)^n - (4/(z - 1)) * Σ(n = 0 to ∞) (-1/(z - 1))^n]
via geometric series
= (1/4) [2/(z - 1) + Σ(n = 0 to ∞) (-1/2)^n (z - 1)^n - Σ(n = 0 to ∞) 4(-1)^n/(z - 1)^(n+1)].

This converges when both |-(z - 1)/2| < 1 and |-1/(z - 1)| < 1
==> 1 < |z - 1| < 2.

I hope this helps!