Fins all ordered triples $ \left(a,b,c\right)$ of positive integers such that $ abc \plus{} ab \plus{} c \equal{} a^3$.
We need to find all ordered triples \((a, b, c)\) of positive integers such that \(abc + ab + c = a^3\).

1. **Rewrite the given equation:**
   \[
   abc + ab + c = a^3
   \]
   Factor out \(c\) on the left-hand side:
   \[
   c(ab + 1) + ab = a^3
   \]
   Rearrange to isolate \(c\):
   \[
   c(ab + 1) = a^3 - ab
   \]
   \[
   c = \frac{a^3 - ab}{ab + 1}
   \]

2. **Analyze the divisibility:**
   Since \(c\) must be an integer, \(\frac{a^3 - ab}{ab + 1}\) must be an integer. This implies that \(ab + 1\) must divide \(a^3 - ab\).

3. **Simplify the expression:**
   \[
   a^3 - ab = a(ab + 1) - ab - a = a^2b + a - ab = a^2b - ab + a = ab(a - 1) + a
   \]
   Therefore, we need \(ab + 1\) to divide \(a(ab - 1) + a\).

4. **Consider the factorization:**
   \[
   a(a^2 - b) = c(1 + ab)
   \]
   Since \(\gcd(a, ab + 1) = 1\), \(a\) must divide \(c\). Let \(c = ak\) for some positive integer \(k\).

5. **Substitute \(c = ak\):**
   \[
   a(a^2 - b) = ak(1 + ab)
   \]
   Simplify:
   \[
   a^3 - ab = ak + abk
   \]
   \[
   a^3 + 1 = abk + ab + ak + 1
   \]
   \[
   a^3 + 1 = (ak + 1)(ab + 1)
   \]

6. **Use the Four Number Theorem:**
   We have \(x, y, z, t\) such that \((x, z) = 1\) and:
   \[
   a + 1 = xy, \quad a^2 - a + 1 = zt, \quad ak + 1 = zy, \quad ab + 1 = xt
   \]
   Since \(k, b < a\), we have:
   \[
   zy \leq a(a - 1) + 1 = zt \quad \text{so} \quad y \leq t
   \]
   \[
   xt \leq (a - 1)a + 1 = zt \quad \text{so} \quad x \leq z
   \]
   If \(z > x\) and \(t > y\), then \(k, b < a - 1\), so:
   \[
   ak + 1, ab + 1 \leq (a - 1)^2 = (xy - 2)^2 < (xy)^2
   \]
   \[
   xyzt < (xy)^2 \quad \text{so} \quad zt < xy \quad \text{contradiction!}
   \]
   Therefore, \(z = x\) or \(t = y\).

7. **Case 1: \(z = x\):**
   \[
   b = a - 1 \quad \text{and} \quad ak + 1 = zy = xy = a + 1 \quad \text{so} \quad k = 1
   \]
   \[
   c = a \quad \text{so} \quad (a, b, c) = (a, a - 1, a)
   \]

8. **Case 2: \(t = y\):**
   \[
   k = a - 1 \quad \text{so} \quad c = a(a - 1)
   \]
   \[
   ab + 1 = xt = xy = a + 1 \quad \text{so} \quad b = 1
   \]
   \[
   (a, b, c) = (a, 1, a(a - 1))
   \]

Conclusion:
The ordered triples \((a, b, c)\) that satisfy the equation are:
\[
(a, a - 1, a) \quad \text{and} \quad (a, 1, a(a - 1))
\]

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