The roots of the equation 
\[ x^3-3ax^2+bx+18c=0 \]
form a non-constant arithmetic progression and the roots of the equation 
\[ x^3+bx^2+x-c^3=0 \]
form a non-constant geometric progression. Given that $a,b,c$ are real numbers, find all positive integral values $a$ and $b$.
We are given two polynomial equations with roots forming arithmetic and geometric progressions, respectively. We need to find all positive integral values of \(a\) and \(b\).

### Equation 1: \(x^3 - 3ax^2 + bx + 18c = 0\)
The roots of this equation form a non-constant arithmetic progression. Let the roots be \(\alpha\), \(\beta\), and \(\gamma\). Since the roots are in arithmetic progression, we can write:
\[
\beta = \alpha + d \quad \text{and} \quad \gamma = \alpha + 2d
\]
where \(d\) is the common difference.

1. **Sum of the roots**:
   \[
   \alpha + \beta + \gamma = 3a
   \]
   Substituting the roots:
   \[
   \alpha + (\alpha + d) + (\alpha + 2d) = 3a \implies 3\alpha + 3d = 3a \implies \alpha + d = a
   \]
   Therefore, \(\beta = a\).

2. **Sum of the product of the roots taken two at a time**:
   \[
   \alpha\beta + \beta\gamma + \gamma\alpha = b
   \]
   Substituting the roots:
   \[
   \alpha(a) + a(a + d) + (a - d)(a) = b \implies a^2 - ad + a^2 + ad + a^2 - d^2 = b \implies 3a^2 - d^2 = b
   \]

3. **Product of the roots**:
   \[
   \alpha\beta\gamma = -18c
   \]
   Substituting the roots:
   \[
   (a - d)a(a + d) = -18c \implies a(a^2 - d^2) = -18c
   \]

### Equation 2: \(x^3 + bx^2 + x - c^3 = 0\)
The roots of this equation form a non-constant geometric progression. Let the roots be \(u\), \(v\), and \(w\). Since the roots are in geometric progression, we can write:
\[
u = \frac{v}{r}, \quad v = v, \quad w = vr
\]
where \(r\) is the common ratio.

1. **Sum of the roots**:
   \[
   u + v + w = -b
   \]
   Substituting the roots:
   \[
   \frac{v}{r} + v + vr = -b \implies v\left(\frac{1}{r} + 1 + r\right) = -b
   \]

2. **Sum of the product of the roots taken two at a time**:
   \[
   uv + vw + wu = 1
   \]
   Substituting the roots:
   \[
   v\left(\frac{v}{r}\right) + v(vr) + \left(\frac{v}{r}\right)(vr) = 1 \implies v^2\left(\frac{1}{r} + r + 1\right) = 1
   \]
   From the above two equations, we get:
   \[
   \frac{1}{v} = -b \implies v = -\frac{1}{b}
   \]

3. **Product of the roots**:
   \[
   uvw = -c^3
   \]
   Substituting the roots:
   \[
   \left(\frac{v}{r}\right)v(vr) = -c^3 \implies v^3 = -c^3 \implies v = -c
   \]
   Since \(v = -\frac{1}{b}\), we have:
   \[
   -\frac{1}{b} = -c \implies c = \frac{1}{b}
   \]

### Combining the Results:
From Equation 1, we have:
\[
a(a^2 - d^2) = -18c
\]
From Equation 2, we have:
\[
bc = 1
\]
Substituting \(c = \frac{1}{b}\) into the first equation:
\[
a(a^2 - d^2) = -18\left(\frac{1}{b}\right) \implies ab(a^2 - d^2) = -18
\]
We also have:
\[
b = 3a^2 - d^2
\]
Substituting \(b\) into the equation:
\[
ab(3a^2 - d^2 - 2a^2) = 18 \implies ab(a^2 - d^2) = 18
\]
Thus, we need to solve:
\[
ab(3a^2 - d^2) = 18
\]
Testing positive integral values, we find:
\[
(a, b) = (2, 9)
\]

The final answer is \( \boxed{ (a, b) = (2, 9) } \)