The difference between two positive numbers is 4 and the difference between their cubes is 316. Find:
their product

Answer

To solve the problem, we will follow these steps:

### Step 1: Define the Variables
Let the two positive numbers be \( x \) and \( y \).

### Step 2: Set Up the Equations
From the problem, we know:
1. The difference between the two numbers:
\[ x - y = 4 \quad \text{(Equation 1)} \]
2. The difference between their cubes:
\[ x^3 - y^3 = 316 \quad \text{(Equation 2)} \]

### Step 3: Use the Identity for Difference of Cubes
We can use the identity for the difference of cubes:
\[ x^3 - y^3 = (x - y)(x^2 + y^2 + xy) \]
Substituting Equation 1 into this identity:
\[ x^3 - y^3 = (4)(x^2 + y^2 + xy) = 316 \]
This simplifies to:
\[ x^2 + y^2 + xy = \frac{316}{4} = 79 \quad \text{(Equation 3)} \]

### Step 4: Express \( x^2 + y^2 \) in Terms of \( xy \)
From Equation 1, we can square both sides:
\[ (x - y)^2 = 4^2 \implies x^2 - 2xy + y^2 = 16 \]
This can be rearranged to express \( x^2 + y^2 \):
\[ x^2 + y^2 = 16 + 2xy \quad \text{(Equation 4)} \]

### Step 5: Substitute Equation 4 into Equation 3
Now, we substitute Equation 4 into Equation 3:
\[ (16 + 2xy) + xy = 79 \]
This simplifies to:
\[ 16 + 3xy = 79 \]

### Step 6: Solve for \( xy \)
Now, isolate \( xy \):
\[ 3xy = 79 - 16 \]
\[ 3xy = 63 \]
\[ xy = \frac{63}{3} = 21 \]

### Conclusion
The product of the two numbers \( x \) and \( y \) is:
\[ \boxed{21} \]