11.5. A cube of size $n \times n \times n$, where $n$ is a natural number, was cut into 99 smaller cubes, of which exactly one has an edge length different from 1 (each of the others has an edge length of 1). Find the volume of the original cube.
Answer: 125.

Let $m$ be the length of the edge of the cube, different from the unit cube. We get the equation $n^{3}-m^{3}=98$ (in natural numbers). Further, $(n-m)\left(n^{2}+n m+m^{2}\right)=98$. The numbers $m$ and $n$ are of the same parity, otherwise $n^{3}-m^{3}$ would be an odd number. Moreover, if $n$ and $m$ are even numbers, then $n^{3}-m^{3}$ would be divisible by 8, but 98 is not divisible by 8. Therefore, $m$ and $n$ are odd, and the first factor of the product $(n-m)\left(n^{2}+n m+m^{2}\right)$ is an even number, while the second is odd. Finally, the second factor is greater than the first: $n^{2}+n m+m^{2}>n>n-m$. From this and the factorization $98=2 \cdot 7 \cdot 7$ into prime factors, we obtain the system of equations $n-m=2, n^{2}+n m+m^{2}=49$. The only solution to this system in natural numbers is the pair $n=5, m=3$.

Remark. The correct answer with verification is worth up to 3 points.