8.2. The product of two natural numbers $a$ and $b$ is a three-digit number, which is the cube of some natural number $k$. The quotient of the numbers $a$ and $b$ is the square of this same number $k$. Find $a, b$, and $k$.
Solution: Let's write the conditions as follows: $a \cdot b=k^{3}, a / b=k^{2}$. Dividing the first by the second, we get $b^{2}=k$. Since $k^{3}$ is a three-digit number, then $b^{6}$ is also a three-digit number. Clearly, $b$ cannot be 1 or 2, since $1^{6}=1, 2^{6}=64$. Next, $b$ can be 3, since $3^{6}=729$. Finally, if $b \geqslant 4$, then $b^{6} \geqslant 4^{6}=16^{3}>10^{3}=1000$, which also does not fit.

Thus, $b=3$. From this, $k=9$ and $a=243$.

Criteria: Only the answer or the answer with verification - 1 point.

The scheme of enumeration, if the enumeration is not complete - no more than 3 points.

During the olympiad, the conditions were distributed with a typo: it was required to find all integer $a$ and $b$, which added another solution $a=-243, b=-3, k=9$. Points were not deducted for the absence of this case.