Cube of Prime is Multiplicatively Perfect

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Let $n \in \Z_{>0}$ be a positive integer which is the cube of a prime number.

Then $n$ is multiplicatively perfect.


Let $n = p^3$ where $p$ is prime.

From Cube of Prime has 4 Positive Divisors, the positive divisors of $n$ are:

$1, p, p^2, p^3$

Thus the product of all the divisors of $n$ is:

$1 \times p \times p^2 \times p^3 = p^6 = n^2$

Hence the result, by definition of multiplicatively perfect.