If you assume abc is the smallest number of each triplet, the solutions
are a=2,b=4,c=3 and a=5,b=6,c=7. The former solution is the one that
yields a sixth power. Pretty clever. I solved this one with a 
combination of a hand calculator, logical reasoning, and brute effort.
Darrel C Jones

abc=243 or 567
I like ur page.
Nils Mellech

Well, not really.  It seems to me that the number of solutions must be a multiple of 3, since if abc is a solution then so must be bca and cab.  But I understand what you meant.
Anyhow, abc = 243 works (243×432×324 = 34,012,224 = 5832^2 = 18^6) as does abc = 567 (567×675×756 = 289,340,100 = 17010^2).
Al Zimmermann

The only two solutions to Robert Reid's problem ("find a three-digit 
number abc such that abc*bca*cab is a square number") up to rotation are
234 * 342 * 423 = 34012224 = 5832^2  ( = 324^3 = 18^6)
576 * 765 * 657 = 289340100 = 17010^2
Considering higher powers (not just squares) yields, in addition to the 
solutions above and the obvious cases a=b=c (abc * bca * cab = aaa^3), 
one additional solution:
468 * 684 * 846 = 272097792 = 648^3
Those are the only solutions to abc * bca * cab = N^k for k > 1.
I also looked at numbers with more digits, such as abcd*bcda*cdab*dabc 
and abcde*bcdea*cdeab*deabc*eabcd, but these did not yield solutions other than the obvious ones:
Michael Brundage

{ abc , bca , cab : abc X bca X cba is a perfect square }
= { 243 , 432 , 324 }  ( Square 34012224 = 5832 ^2 )
= { 567 , 675 , 756 }  ( Square 289340100 = 17010 ^2 )
Sudipta Das

Given the language of your original challenge:
"Find a three digit number abc such that abc×bca×cab is a square 
number."  there are, pedantically speaking, six solutions:
abc = 432,  abc = 324,  abc = 243,  abc = 567,  abc = 675,  abc = 756
However there are only two unique squares. The 432, et.al. group are 
the ones that generate the sixth power, who's root is 18.
Thanks for the fun challenge!
Mike Kurtinitis

Robert Reid's three digit number is:
1:  243x432x324 = 34012224 = 5823^2
2:  567x675x756 = 289340100 = 17010^2
--Koshi Arai

Solved using Perl
243 x 324 x 432 = 5832^2
567 x 675 x 756 = 17010^2
#!/usr/bin/perl
use warnings;
use strict;

my ($a, $b, $c);
my ($p, $q, $r);
my ($t);

for (my $x=100; $x<1000; $x++) {
    $a=substr($x,0,1);
    $b=substr($x,1,1);
    $c=substr($x,2,1);
    $p=int($a.$b.$c);
    $q=int($b.$c.$a);
    $r=int($c.$a.$b);
    $t=sqrt($p*$q*$r);
    if (length($t) eq length(int($t))) {
        print "$a $b $c", "\n";
    }
}
exit;
Michael J D Dufour  BSc. (Hon.) MSc.

Nice to have an easy one for once!
243 x 432 x 324 = 5832^2 = 18^6
567 x 675 x 756 = 17010^2
John Gowland

The answer for abc are 243 and 567.
abc=243 gives abc*bca*cab as 6th power i.e. 18^6
Shyam Sunder Gupta

1. 243
2. 567
Jared Marks
James Lewis Melby
Matt Coury

I solve a lot of puzzles from this site and I enjoy almost every 
puzzle. 
But I don't send the answers. A friend of mine suggested that I should 
send them, so I start with this easier one. This site is Really a great 
work and thanks a lot for each and every puzzle that I enjoyed.
1. abc = 243 (243 x 432 x 324 = 5832 squared = 18^6)
2. abc = 567 (567 x 675 x 756 = 17010 squared)
Anupam

234 x 243 x 432 = 34012224 = 5832 x 5832
567 x 576 x 756 = 289340100 = 17010 x 17010
by Javier Arbonés (Barcelona-Spain)