To find the residues of the function 

\[
f(z) = \frac{\log(z)}{z^3 + 1}
\]

we need to follow these steps:

### 1. **Find the poles of the function**:
   The residues are located at the poles of the function. The poles occur when the denominator is zero, i.e., when:

   \[
   z^3 + 1 = 0
   \]

   Solving for \(z\), we get:

   \[
   z^3 = -1
   \]

   The solutions to this equation are the **cube roots of -1**. These roots are:

   \[
   z = e^{i\pi/3}, \quad z = e^{i\pi}, \quad z = e^{i5\pi/3}
   \]

   More explicitly, the poles are:

   \[
   z_1 = -1, \quad z_2 = e^{i2\pi/3}, \quad z_3 = e^{i4\pi/3}
   \]

### 2. **Classify the poles**:
   Since \(z^3 + 1 = 0\) is a cubic equation, each of these poles is a **simple pole**. We can see this by noting that the derivative of \(z^3 + 1\) is non-zero at these points.

### 3. **Find the residues**:
   To compute the residues at each pole \(z_k\), we use the formula for the residue of a simple pole:

   \[
   \text{Res}(f, z_k) = \lim_{z \to z_k} (z - z_k) f(z)
   \]

   That is, we need to evaluate:

   \[
   \text{Res}(f, z_k) = \lim_{z \to z_k} \frac{(z - z_k)\log(z)}{z^3 + 1}
   \]

   The function \(z^3 + 1\) can be factored as:

   \[
   z^3 + 1 = (z - z_k)(z - z_2)(z - z_3)
   \]

   Thus, the residue at each \(z_k\) simplifies to:

   \[
   \text{Res}(f, z_k) = \frac{\log(z_k)}{3z_k^2}
   \]

   Now, compute the residues at each pole:

   - **At \(z_1 = -1\)**:
     \[
     \text{Res}(f, -1) = \frac{\log(-1)}{3(-1)^2} = \frac{i\pi}{3}
     \]

   - **At \(z_2 = e^{i2\pi/3}\)**:
     \[
     \text{Res}(f, e^{i2\pi/3}) = \frac{\log(e^{i2\pi/3})}{3(e^{i2\pi/3})^2} = \frac{i2\pi/3}{3e^{i4\pi/3}}
     \]
     Simplifying, we get:
     \[
     \text{Res}(f, e^{i2\pi/3}) = \frac{i2\pi/9}{e^{i4\pi/3}}
     \]

   - **At \(z_3 = e^{i4\pi/3}\)**:
     \[
     \text{Res}(f, e^{i4\pi/3}) = \frac{\log(e^{i4\pi/3})}{3(e^{i4\pi/3})^2} = \frac{i4\pi/3}{3e^{i8\pi/3}}
     \]
     Simplifying further:
     \[
     \text{Res}(f, e^{i4\pi/3}) = \frac{i4\pi/9}{e^{i8\pi/3}}
     \]

### Conclusion:
The residues of \( \frac{\log(z)}{z^3 + 1} \) at the poles \(z_1 = -1\), \(z_2 = e^{i2\pi/3}\), and \(z_3 = e^{i4\pi/3}\) are:

- \( \text{Res}(f, -1) = \frac{i\pi}{3} \)
- \( \text{Res}(f, e^{i2\pi/3}) = \frac{i2\pi/9}{e^{i4\pi/3}} \)
- \( \text{Res}(f, e^{i4\pi/3}) = \frac{i4\pi/9}{e^{i8\pi/3}} \)

Would you like further details or clarification on any steps?

### Related Questions:
1. How do you compute residues for higher-order poles?
2. What is the importance of the residue theorem in complex analysis?
3. How does the logarithmic branch cut affect the evaluation of residues?
4. Can we apply contour integration to find the sum of residues?
5. What are the applications of computing residues in physics?

### Tip:
When computing residues at roots of unity, consider using polar coordinates for simplifications involving logarithms and powers of complex numbers.

Explore how to find the residues of the function log(z) / (z^3 + 1) by identifying the poles, classifying them, and calculating the residues at each pole using the residue formula. This problem covers key concepts in complex analysis and is suitable for undergraduate students.

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