Answer on Abstract Algebra Question for Josh Haver
n= p1 * p2 * … * pk,
be such a presentation. Some of multiples may repeat.
Then n3 has the following presentation
n3= p13 * p23 * … * pk3=
=p1*p1*p1*p2*p2*p2* … *
Hence if p divides n3, then it divides n, and therefore p3 divides n3.
Similarly for q.
Since p and q are distinct numbers, it follows that
(pq)3 = p3 q3
also divides n3
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