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 =

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**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.

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**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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