Two positive numbers x and y are such that `x gt y`. If the difference of these numbers is 5 and their product is 24, find:
difference of their cubes

Answer

To solve the problem step by step, we need to find the two positive numbers \( x \) and \( y \) such that \( x > y \), their difference is 5, and their product is 24. We will then find the difference of their cubes.

### Step 1: Set up the equations
We know from the problem statement:
1. \( x - y = 5 \) (Equation 1)
2. \( x \cdot y = 24 \) (Equation 2)

### Step 2: Express \( x \) in terms of \( y \)
From Equation 1, we can express \( x \) in terms of \( y \):
\[ x = y + 5 \]

### Step 3: Substitute \( x \) in Equation 2
Now we substitute \( x \) in Equation 2:
\[ (y + 5) \cdot y = 24 \]
Expanding this gives:
\[ y^2 + 5y = 24 \]

### Step 4: Rearrange the equation
Rearranging the equation to set it to zero:
\[ y^2 + 5y - 24 = 0 \]

### Step 5: Factor the quadratic equation
Next, we need to factor the quadratic equation. We are looking for two numbers that multiply to \(-24\) and add to \(5\). The numbers \(8\) and \(-3\) satisfy this:
\[ (y + 8)(y - 3) = 0 \]

### Step 6: Solve for \( y \)
Setting each factor to zero gives us:
1. \( y + 8 = 0 \) → \( y = -8 \) (not valid since \( y \) must be positive)
2. \( y - 3 = 0 \) → \( y = 3 \)

### Step 7: Find \( x \)
Now, substituting \( y = 3 \) back into the equation for \( x \):
\[ x = y + 5 = 3 + 5 = 8 \]

### Step 8: Find the difference of their cubes
Now we need to calculate the difference of their cubes:
\[ x^3 - y^3 = 8^3 - 3^3 \]
Calculating the cubes:
\[ 8^3 = 512 \quad \text{and} \quad 3^3 = 27 \]
Thus,
\[ x^3 - y^3 = 512 - 27 = 485 \]

### Final Answer
The difference of their cubes is:
\[ \boxed{485} \]