<s>[INST] Concerning the product of all unique positive divisors 
If the product of all the unique positive divisors of $n$, a positive integer which is not a perfect cube, is $n^2$, then the product of all the unique positive divisors of $n^2$ is:
  (A)    $n^3$
  (B)    $n^4$
  (C)    $n^6$
  (D)    $n^8$
  (E)    $n^9$

I tried using the formula: $\frac{n^{d(n)}}{2}$, where $d(n)$ is the number of factors and I got the answer as $n^7/2$.
 [/INST] Since the product of positive divisors of $n$ is $n^{\frac{d(n)}{2}}$. We want 
$$n^{\frac{d(n)}{2}}=n^2.$$
This means $d(n)=4$. So either $n=p^3$ or $n=pq$. The first possibility is ruled out since $n$ is not a cube. Thus $n=pq$, where $p$ and $q$ are distinct primes. Then $d(n^2)=(2+1)(2+1)=9$ and the product of divisors of $n^2$ is 
$$(n^2)^{\frac{d(n^2)}{2}}=(n^2)^{9/2}=n^9$$ 
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