Find the smallest positive value taken by $a^3 + b^3 + c^3 - 3abc$ for positive integers $a$, $b$, $c$ .
 Find all $a$, $b$, $c$ which give the smallest value
1. We start with the expression \(a^3 + b^3 + c^3 - 3abc\). We can use the identity:
   \[
   a^3 + b^3 + c^3 - 3abc = \frac{1}{2}(a+b+c)\left[(a-b)^2 + (b-c)^2 + (c-a)^2\right]
   \]
   This identity can be derived from the factorization of the polynomial \(a^3 + b^3 + c^3 - 3abc\).

2. To find the smallest positive value of \(a^3 + b^3 + c^3 - 3abc\) for positive integers \(a\), \(b\), and \(c\), we need to analyze the expression \(\frac{1}{2}(a+b+c)\left[(a-b)^2 + (b-c)^2 + (c-a)^2\right]\).

3. Notice that \(a\), \(b\), and \(c\) are positive integers, so \(a + b + c \geq 3\). If \(a = b = c = 1\), then:
   \[
   a^3 + b^3 + c^3 - 3abc = 1^3 + 1^3 + 1^3 - 3 \cdot 1 \cdot 1 \cdot 1 = 3 - 3 = 0
   \]
   However, we are looking for the smallest positive value, so we need \(a + b + c > 3\).

4. Next, consider the case where \(a + b + c = 4\). We need to find \(a\), \(b\), and \(c\) such that \((a-b)^2 + (b-c)^2 + (c-a)^2\) is minimized but not zero. The smallest non-zero value for \((a-b)^2 + (b-c)^2 + (c-a)^2\) is 2, which occurs when two of the numbers are equal and the third is different by 1.

5. Let's try \(a = 1\), \(b = 1\), and \(c = 2\):
   \[
   a + b + c = 1 + 1 + 2 = 4
   \]
   \[
   (a-b)^2 + (b-c)^2 + (c-a)^2 = (1-1)^2 + (1-2)^2 + (2-1)^2 = 0 + 1 + 1 = 2
   \]
   Substituting these values into the expression:
   \[
   a^3 + b^3 + c^3 - 3abc = 1^3 + 1^3 + 2^3 - 3 \cdot 1 \cdot 1 \cdot 2 = 1 + 1 + 8 - 6 = 4
   \]

6. Therefore, the smallest positive value of \(a^3 + b^3 + c^3 - 3abc\) is 4, and it is achieved when \(a = 1\), \(b = 1\), and \(c = 2\).

The final answer is \(\boxed{4}\)