$\left.\begin{array}{l}{[\text { Decimal numeral system }]} \\ {\left[\begin{array}{l}\text { Case analysis }\end{array}\right]} \\ {[\underline{\text { Divisibility rules for 11 }}]} \\ {[\underline{\text { Divisibility rules for 3 and 9 }}]}\end{array}\right]$

A two-digit number is appended to the right of a certain natural number, and then another two-digit number is appended. The resulting number turns out to be equal to the cube of the sum of the three original numbers. Find all possible triples of the original numbers.
Let $a$ denote the first natural number, and $b$ and $c$ the two-digit numbers written after it. Let $x = a + b + c$. By the condition,

$10^{4} a + 100 b + c = x^{3}$.

If $x \geq 100$, then $x^{3} \geq 10^{4} x = 10^{4}(a + b + c) > 10^{4} a + 100 b + c$, which means the equation has no solutions.

Therefore, $x$ is a two-digit number, $a$ is either a single-digit or a two-digit number, and $x^{3}$ is a five- or six-digit number.

Notice that the number $x^{3} - x = 9999 a + 99 b$ is divisible by 99. Since $x^{3} - x = x(x-1)(x+1)$, one of the numbers $x-1$, $x$, $x+1$ is divisible by 9 and another by 11. Since $22 \leq x \leq 99$, the following cases are possible:

1) $x = 44 (x+1 = 45), 44^{3} = 85184, 8 + 51 + 84 > 44$
2) $x = 45 (x-1 = 44), 45^{3} = 91125, a = 9, b = 11, c = 25$
3) $x = 54 (x+1 = 55), 54^{3} = 157464, 15 + 74 + 64 > 54$
4) $x = 55 (x-1 = 54), 55^{3} = 166375, 16 + 63 + 75 > 55$
5) $x = 89 (x-1 = 88, x+1 = 90), 89^{3} = 704969, 70 + 49 + 69 > 89$;
6) $x = 98 (x+1 = 99), 98^{3} = 941192, 94 + 11 + 92 > 98$
7) $x = 99, x^{3} = 970299, 2$ is not a two-digit number.

## Answer

$9, 11, 25$.