1. **Stating the problem:** We have three consecutive positive integers, and $n$ is the middle integer. We multiply these three integers, then add $n$ to the product, and we want to prove that this sum is a cube number. 2. **Expressing the integers:** Let the three consecutive integers be $n-1$, $n$, and $n+1$. 3. **Multiplying the integers:** The product of the three integers is: $$ (n-1) \times n \times (n+1) $$ 4. **Simplifying the product:** This is a product of three consecutive integers which can be rewritten using the difference of squares: $$ (n-1)(n+1) = n^2 - 1 $$ Hence the product is: $$ n(n^2 - 1) = n^3 - n $$ 5. **Adding $n$ to the product:** According to the problem, we add $n$ to the product: $$ (n^3 - n) + n = n^3 $$ 6. **Conclusion:** The result simplifies to $n^3$, which is a cube number, specifically the cube of the middle integer $n$. Hence proven that the sum of the product of the three consecutive integers and the middle integer itself is a cube number. **Final answer:** The result is exactly $n^3$, a perfect cube.