Determine all pairs of integers $(a, b)$  that solve the equation $a^3 + b^3 + 3ab = 1$.
To determine all pairs of integers \((a, b)\) that solve the equation \(a^3 + b^3 + 3ab = 1\), we can use the following lemma:

**Lemma:** If \(x\), \(y\), and \(z\) are real numbers such that \(x^3 + y^3 + z^3 = 3xyz\), then \(x + y + z = 0\) or \(x = y = z\).

Given the equation:
\[ a^3 + b^3 + 3ab = 1 \]

We can rewrite it as:
\[ a^3 + b^3 + (-1)^3 = 3ab(-1) \]

This matches the form of the lemma with \(x = a\), \(y = b\), and \(z = -1\). According to the lemma, we have two cases to consider:

1. \(a + b - 1 = 0\)
2. \(a = b = -1\)

### Case 1: \(a + b - 1 = 0\)
This simplifies to:
\[ a + b = 1 \]

For any integer \(a\), we can find \(b\) as:
\[ b = 1 - a \]

Thus, the pairs \((a, b)\) that satisfy this case are:
\[ (a, 1 - a) \]
for any integer \(a\).

### Case 2: \(a = b = -1\)
In this case, the pair is:
\[ (-1, -1) \]

### Verification:
We need to verify that these pairs indeed satisfy the original equation.

1. For \((a, 1 - a)\):
   \[ a^3 + (1 - a)^3 + 3a(1 - a) = 1 \]
   Expanding and simplifying:
   \[ a^3 + (1 - 3a + 3a^2 - a^3) + 3a - 3a^2 = 1 \]
   \[ a^3 + 1 - 3a + 3a^2 - a^3 + 3a - 3a^2 = 1 \]
   \[ 1 = 1 \]
   This is true for any integer \(a\).

2. For \((-1, -1)\):
   \[ (-1)^3 + (-1)^3 + 3(-1)(-1) = 1 \]
   \[ -1 - 1 + 3 = 1 \]
   \[ 1 = 1 \]
   This is also true.

Therefore, the pairs \((a, 1 - a)\) for any integer \(a\) and \((-1, -1)\) are the solutions.

The final answer is \( \boxed{ (a, 1 - a) } \) for any integer \(a\) and \((-1, -1)\).