The first plug in trail and error approach does not make sense as one has to depend on luck rather than an approach to solve the problem in a systematic way.

In the second approach the solution says:
we can figure out that n must be the product of two distinct primes, say p and q (so that n = pq)
I am not sure how to conclude that n must be the product of two distinct primes.


If the product of all the unique positive divisors of n, a positive integer which is not a perfect cube, is n2, then the product of all the unique positive divisors of n2 is
(A) n3
(B) n4
(C) n6
(D) n8
(E) n9

Solution:
The first task is to figure out a positive integer n for which the given condition is true: that all the unique positive divisors multiply up to n2.

Let’s first take a number-testing approach.
If n = 1, the condition holds, but we can quickly see that all the answer choices are equal in this case.
If n = 2, the condition doesn’t hold. Take the unique positive factors: 1 and 2. The product is 2, not 22.
The same is true for n = 3 or for any other prime number.
If n = 4, the condition still doesn’t hold. The unique positive divisors are 1, 2, and 4, which multiply up to 8, not 16.
If n = 6, the condition holds (thankfully). The unique positive divisors are 1, 2, 3, and 6, which multiply up to 36, or 62.

Thus, we can solve the problem if we figure out the product of all the unique positive divisors of 36. We don’t need the digits; we just need to know what that number is as a power of 6 (our n). The fast, organized way is to take the factors in pairs:
1 × 36 = 36 (=62)
2 × 18 = 36 (=62)
3 × 12 = 36 (=62)
4 × 9 = 36 (=62)
And finally there’s a 6, of course, the square root of 36. We only count this once (the question specifies unique divisors).
The product of the four “6-squared’s” (62) and the final 6 is 62 × 62 × 62 × 62 × 6 = 69.

This gives us the correct answer of E.

Alternatively, we could solve this problem in a more abstract way. First, we can figure out that n must be the product of two distinct primes, say p and q (so that n = pq). The only unique positive factors of n would then be 1, p, q, and pq (that is, n itself). Then we could write n2 as p2q2. (By the way, if n were allowed to be a cube of a prime number (such that n = p3), then the product of all of n's factors would also be n2, but this possibility is explicitly ruled out.)
In terms of p and q, then, the unique factors of n2 would then be as follows:

1, p, p2
q, qp, qp2
q2, q2p, q2p2

The product of all these factors is q9p9, or n9.

The correct answer is E.

what "doesn't make sense" about that?

there are lots and lots and lots (and lots) of GMAT problems that require trial-and-error solutions.

this is an intentional component of the test. the test writers want to make sure that people whose ONLY approaches are systematic DO NOT score in the highest ranks!
the test is purposely designed to reward mental flexibility and to discourage mental rigidity.

(here's another problem on which you *MUST* use trial and error; search for another thread if you want to discuss it:
The number 75 can be written as the sum of the squares of 3 different positive integers. What is the sum of these 3 integers?
A. 17
B. 16
C. 15
D. 14
E. 13 )



if there are three or more primes p,q,r, then you already have the following factors: p, q, r, pq, pr, qr, pqr. so your product is already equal to n^3, i.e., too big.

that's really difficult to come up with, of course.

if your instinct, at this point, is *still* to ignore the plug-in method and concentrate 100% on textbook methods, then you will find it VERY difficult to improve your quant performance.
if you open your mind up to "alternative" methods of solution, you will suddenly find the quant portion of the test much more friendly.