The sum of the squares of three positive numbers is $160$. One of the numbers is equal to the sum of the other two. The difference between the smaller two numbers is $4.$ What is the difference between the cubes of the smaller two numbers?

[i]Author: Ray Li[/i]

[hide="Clarification"]The problem should ask for the positive difference.[/hide]
1. Let the three positive numbers be \(a\), \(b\), and \(c\), with \(a\) being the largest. We are given the following conditions:
   \[
   a^2 + b^2 + c^2 = 160
   \]
   \[
   a = b + c
   \]
   \[
   b - c = 4
   \]

2. Substitute \(a = b + c\) into the first equation:
   \[
   (b + c)^2 + b^2 + c^2 = 160
   \]
   Expanding \((b + c)^2\):
   \[
   b^2 + 2bc + c^2 + b^2 + c^2 = 160
   \]
   Combine like terms:
   \[
   2b^2 + 2c^2 + 2bc = 160
   \]
   Divide the entire equation by 2:
   \[
   b^2 + c^2 + bc = 80
   \]

3. We also have the equation \(b - c = 4\). We need to find the difference between the cubes of \(b\) and \(c\). First, solve for \(b\) and \(c\) using the system of equations:
   \[
   b - c = 4
   \]
   \[
   b^2 + c^2 + bc = 80
   \]

4. Express \(b\) in terms of \(c\):
   \[
   b = c + 4
   \]

5. Substitute \(b = c + 4\) into the equation \(b^2 + c^2 + bc = 80\):
   \[
   (c + 4)^2 + c^2 + (c + 4)c = 80
   \]
   Expand and simplify:
   \[
   c^2 + 8c + 16 + c^2 + c^2 + 4c = 80
   \]
   Combine like terms:
   \[
   3c^2 + 12c + 16 = 80
   \]
   Subtract 80 from both sides:
   \[
   3c^2 + 12c + 16 - 80 = 0
   \]
   Simplify:
   \[
   3c^2 + 12c - 64 = 0
   \]

6. Solve the quadratic equation using the quadratic formula \(c = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\):
   \[
   c = \frac{-12 \pm \sqrt{12^2 - 4 \cdot 3 \cdot (-64)}}{2 \cdot 3}
   \]
   \[
   c = \frac{-12 \pm \sqrt{144 + 768}}{6}
   \]
   \[
   c = \frac{-12 \pm \sqrt{912}}{6}
   \]
   \[
   c = \frac{-12 \pm 6\sqrt{38}}{6}
   \]
   \[
   c = -2 \pm \sqrt{38}
   \]
   Since \(c\) must be positive:
   \[
   c = -2 + \sqrt{38}
   \]

7. Find \(b\):
   \[
   b = c + 4 = -2 + \sqrt{38} + 4 = 2 + \sqrt{38}
   \]

8. Calculate the difference between the cubes of \(b\) and \(c\):
   \[
   b^3 - c^3 = (2 + \sqrt{38})^3 - (-2 + \sqrt{38})^3
   \]

9. Use the identity for the difference of cubes:
   \[
   b^3 - c^3 = (b - c)(b^2 + bc + c^2)
   \]
   We know \(b - c = 4\) and \(b^2 + bc + c^2 = 80\):
   \[
   b^3 - c^3 = 4 \cdot 80 = 320
   \]

The final answer is \(\boxed{320}\).