Is there a cube number whose decimal representation is of the form ababab1?
Assuming $a \neq 0$, we are looking for a number $x$ whose cube is a seven-digit number, so $x \geq 100$ and $x \leq 215$, since $216^{3}=10077696$ is already an eight-digit number.

If $x^{3}=\overline{a b a b a b 1}$, then $x^{3}-1=\overline{a b a b a b 0}$, which means it is divisible by 10, and since the sum of its digits is $3(a+b)$, it is also divisible by 3. We will show that if $x^{3}-1$ is divisible by 30, then $x-1$ is also divisible by 30.

$x-1$ is divisible by 10, because if $x^{3}$ ends in 1, then, by checking the cubes of the odd single-digit numbers, $x$ must also end in 1.

$x^{3}$ gives a remainder of 1 when divided by 3 if and only if $x$ does. This can be similarly verified by checking the cubes of the remainders (0, 1, and 2). This means that $x-1$ is divisible by 3.

There are only four numbers divisible by 30 between 99 and 214, so $x$ could be 121, 151, 181, and 211. By cubing these, we can see that the only solution to the problem is $211^{3}=9393931$.

Note. Most solvers found the single solution with more or less calculation. Somogyi Dávid (Fazekas F. Főv. Gyak. Gimn., 9th grade) also checked whether it is possible for $a=0$, or for any digit $c$, that a cube number is of the form $\overline{a b a b a b c}$. By writing out the table of the remainders of $x$ and $x^{3}$ when divided by 30, he showed that apart from the above solution, only 0, 1, and 8 are cube numbers whose decimal representation is of the form $\overline{a b a b a b c}$.