
When tackling complex reasoning tasks, you have access to the following actions. Use them as needed to progress through your thought process.

[ASSESS]

[ADVANCE]

[VERIFY]

[SIMPLIFY]

[SYNTHESIZE]

[PIVOT]

[OUTPUT]

You should strictly follow the format below:

[ACTION NAME]

# Your action step 1

# Your action step 2

# Your action step 3

...

Next action: [NEXT ACTION NAME]


system

Given that all the desired numbers are divisible by 6, they must include the prime factors 2 and 3. Let us then set \(a_{1}=2^{m} \cdot 3^{n}\). For the subsequent numbers \(a_{2}, a_{3}, \ldots, a_{9}\), the exponents of 2 monotonically increase while the exponents of 3 monotonically decrease, for example, \(a_{2}=2^{m+1} \cdot 3^{n-1}, a_{3}=2^{m+2} \cdot 3^{n-2}, \ldots, a_{9}=2^{m+8} \cdot 3^{n-8}\). In this way, no number can be a multiple of another, thus satisfying the second condition of the problem.

Next, considering the cubes of these numbers, the smallest exponent for 2 is \(3m\) and it must be no less than the largest exponent for 2 in the squares of the numbers, i.e., \(2(m+8)\). From \(3m \geq 2(m+8)\), we get \(m \geq 16\).

Similarly, for the exponents of 3, the condition is \(3(n-8) \geq 2n\), resulting in \(n \geq 24\). Therefore, with \(m \geq 16\) and \(n \geq 24\), the cube of each of the numbers \(a_{1}, \ldots, a_{9}\) will be divisible by the square of any other.

Present the answer in LaTex format: \boxed{Your answer}
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