To prove that if \( a + b + c = 0 \), then \( a^3 + b^3 + c^3 = 3abc \), we can use the identity for the sum of cubes.

### Step-by-step proof:

1. **Use the identity for the sum of cubes**:

The identity states:
\[
a^3 + b^3 + c^3 - 3abc = (a+b+c)(a^2 + b^2 + c^2 - ab - ac - bc)
\]

2. **Substitute \(a + b + c = 0\)**:

Since \( a + b + c = 0 \), we substitute into the identity:
\[
a^3 + b^3 + c^3 - 3abc = 0 \cdot (a^2 + b^2 + c^2 - ab - ac - bc)
\]
This simplifies to:
\[
a^3 + b^3 + c^3 - 3abc = 0
\]
Therefore, we conclude that:
\[
a^3 + b^3 + c^3 = 3abc
\]

### Conclusion:

We have shown that if \( a + b + c = 0 \), then \( a^3 + b^3 + c^3 = 3abc \). This completes the proof.