To express \(10032\) as a product of identical factors, we first need to factor \(10032\).
Let's start by finding the prime factorization of \(10032\):
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Divide by \(2\) (since \(10032\) is even): \[ 10032 \div 2 = 5016 \]
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Continue dividing by \(2\): \[ 5016 \div 2 = 2508 \] \[ 2508 \div 2 = 1254 \] \[ 1254 \div 2 = 627 \]
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Now \(627\) is not even, so we try dividing by \(3\): \[ 627 \div 3 = 209 \]
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The next number \(209\) needs to be factored:
- It is not divisible by \(2\), \(3\), or \(5\).
- Check divisibility by \(7\): \(209 \div 7 \approx 29.857\) (not divisible).
- Check divisibility by \(11\): \(209 \div 11 \approx 19.909\) (not divisible).
- Check divisibility by \(13\): \(209 \div 13 = 16.0769\) (not divisible).
- Check divisibility by \(19\): \(209 \div 19 = 11\).
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Thus, \(209 = 11 \times 19\).
Putting all these factors together, we have: \[ 10032 = 2^4 \times 3^1 \times 11^1 \times 19^1. \]
Next, to express this as a product of identical factors, we need to look for the largest square or to take pairs out. Since \(2^4\) is a perfect square: \[ 2^4 = (2^2)^2 = 4^2. \] So we can rewrite \(10032\) as: \[ 10032 = (4 \times 3 \times 11^{1/2} \times 19^{1/2})^2. \] Simplifying yields: \[ 10032 = 4^2 \times (3 \times 11^{1/2} \times 19^{1/2})^2. \]
However, this does not yield identical integer factors, and therefore this attempt isn't giving direct identical factors.
It is evident \(10032\) is not a perfect square, and hence we cannot express it as identical integer factors effectively.
But the answer could be nicely phrased in terms of its primes as: \[ 10032 = (2^2 \times 3 \times 11 \times 19) \times 4^2. \] Thus, expressing \(10032\) neatly as the product of cubes: \[ (2^4 \times 3 \times 11 \times 19) = (2^2 \times 3^{1/2} \times 11^{1/2} \times 19^{1/2})^2. \] Giving: \[ = 2^4 \times 3 \times 11 \times 19 = 4^2 \cdot 3 \cdot 11 \cdot 19 \text{ as a square form or simply }. \]
So, my answer simplifies to what you wanted in a product form: \[ 10032 = 4 \times 4 \cdot 3 \cdot 11 \cdot 19 \] for \(n\) being more concise is always left in individual prime factors.
Thus, \(10032\) cannot be expressed perfectly as identical products but rather an arrangement of its factorization, clearly showing: \[ \boxed{10032} \] retains its fullest expression for clarity.