I will explain three methods to find the residues and you decide which one is easier for you to understand.

Let [imath]\displaystyle z_1 = e^{\frac{i\pi}{3}} = \frac{1}{2} + \text{i}\frac{\sqrt{3}}{2} = \sqrt[3]{-1} \ \ \ \ \ [/imath] (Our main zero.)

*First method.*
It is exactly what professor fresh_42 wrote in post #46. We have a Laurent series [imath]\displaystyle f(z) = \sum_{k=-\infty}^{\infty}a_k(z-z_0)^k[/imath]

And we want to use it to expand our function [imath]\displaystyle \frac{1}{z^3 + 1} = \left(\frac{1}{z + 1}\right)\left(\frac{1}{z - z_1}\right)\left(\frac{1}{z - z_2}\right)[/imath]

where [imath]z_2 = e^{\frac{i5\pi}{3}}[/imath].

Since [imath]\displaystyle \frac{1}{z + 1}[/imath] is analytic at [imath]\displaystyle z_1[/imath], it has a Taylor expansion equal to:

[imath]\displaystyle \frac{1}{1 + \sqrt[3]{-1}} - \frac{z - \sqrt[3]{-1}}{(1 + \sqrt[3]{-1})^2} + \frac{(z - \sqrt[3]{-1})^2}{(1 + \sqrt[3]{-1})^3} - . . . .
[/imath]

Also since [imath]\displaystyle \frac{1}{z - z_2}[/imath] is analytic at [imath]\displaystyle z_1[/imath], it has a Taylor expansion equal to:

[imath]\displaystyle -\frac{\text{i}}{\sqrt{3}} + \frac{1}{3}\left(z - \sqrt[3]{-1}\right) + \frac{\text{i}(z - \sqrt[3]{-1})^2}{3\sqrt{3}} - . . . .[/imath]

We are only interested in the first term from the expansions, so our Laurent expansion is:

[imath]\displaystyle \frac{1}{z^3 + 1} =\left(\frac{1}{1 + \sqrt[3]{-1}}\right)\left(\frac{1}{z + z_1}\right)\left(-\frac{\text{i}}{\sqrt{3}}\right) + ....[/imath]

The residue of [imath]\displaystyle \frac{1}{z^3 + 1}[/imath] at [imath]\displaystyle z_1[/imath] is the coefficient of [imath]\displaystyle \frac{1}{z + z_1}[/imath] in the Laurent expansion which is [imath]\displaystyle \left(\frac{1}{1 + \sqrt[3]{-1}}\right)\left(-\frac{\text{i}}{\sqrt{3}}\right) = \ \ \ ?[/imath]

It is your job to simplify it.

**Second method.**
Let [imath]\displaystyle z_0 = e^{\frac{i\pi}{3}} = \frac{1}{2} + \text{i}\frac{\sqrt{3}}{2} = \sqrt[3]{-1}[/imath]

And

Let [imath]\displaystyle \frac{1}{z^3 + 1} = \frac{p(z)}{q(z)}[/imath]

This means:

[imath]\displaystyle p(z) = 1[/imath]

[imath]\displaystyle q(z) = z^3 + 1[/imath]

[imath]\displaystyle q'(z) = 3z^2[/imath]

If the condition below is satisfied:

[imath]\displaystyle p(z_0) \neq 0[/imath]

[imath]\displaystyle q(z_0) = 0[/imath]

[imath]\displaystyle q'(z_0) \neq 0[/imath]

Then,

The residue of [imath]\displaystyle \frac{1}{z^3 + 1}[/imath] at [imath]\displaystyle z_0[/imath] is [imath]\displaystyle \frac{p(z_0)}{q'(z_0)}[/imath]

Try it

* Third method.*
I think that this is the easiest one. Use this formula:

The residue of [imath]\displaystyle \frac{1}{z^3 + 1}[/imath] at [imath]\displaystyle z_0 = \lim_{z\rightarrow z_0}\frac{1}{(k - 1)!}\frac{d^{k-1}}{dz^{k-1}}[(z - z_0)f(z)][/imath]

where [imath]\displaystyle k[/imath] is the order of the pole

[imath]\displaystyle z_0[/imath] is our zero inside the contour

[imath]\displaystyle \frac{d^{k-1}}{dz^{k-1}}[/imath] is the kth derivative

[imath]\displaystyle f(z)[/imath] is our main function

Examples of the order of the pole for: [imath]z^2 - z + 1[/imath]

[imath]\displaystyle \frac{1}{z^3 + 1} = \frac{1}{(z+1)(z^2 - z + 1)}[/imath], here [imath]\displaystyle k = 1[/imath]

[imath]\displaystyle \frac{1}{(z+1)(z^2 - z + 1)^2}[/imath], here [imath]\displaystyle k = 2[/imath]

We will use the formula to find the residue.

The residue at [imath]\displaystyle z_0 = e^{\frac{i\pi}{3}}[/imath] is [imath]\displaystyle \lim_{z\rightarrow e^{\frac{i\pi}{3}}}\frac{1}{(1 - 1)!}\frac{d^{1 - 1}}{dz^{1 -1}}\left[(z - e^{\frac{i\pi}{3}})\frac{1}{z^3 + 1}\right][/imath]

[imath]\displaystyle = \lim_{z\rightarrow e^{\frac{i\pi}{3}}}\frac{z - e^{\frac{i\pi}{3}}}{z^3 + 1} = \lim_{z\rightarrow e^{\frac{i\pi}{3}}}\frac{1}{3z^2}[/imath]

And it is your job to solve the limit