Posted by Gary on Thursday, July 9, 2009 at 10:50pm.
PLease help with this problem. My friend and I differ on answer.
(3abc^2 )^4/(7a^2 b)^4 =

Math  DrBob222, Thursday, July 9, 2009 at 11:02pm
81c^8/2401a^4.'
There may be a factor for 81 and 2401 but it isn't obvious if there is one.

Math  Gary, Friday, July 10, 2009 at 12:06am
thank you very much, my friend was right I was wrong

Math  drwls, Friday, July 10, 2009 at 7:03am
2401 = 7*7*7*7
81 = 3*3*3*3
There are no common factors
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