To some natural number, two two-digit numbers were appended sequentially to the right. The resulting number turned out to be equal to the cube of the sum of the three original numbers. Find all possible triples of the original numbers.

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Let $a$ denote the first natural number, and $b$ and $c$ the two-digit numbers written after it. Let $x = a + b + c$. According to 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. Additionally, $x \geq 22$ (since $21^{3} = 9261$ - a four-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 $x-1$, $x$, or $x+1$ 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) $x = 99, x^{3} = 970299, 2$ - not a two-digit number.

## Answer

$9, 11, 25$.