PS. can anyone write this using Latex too??
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10 b^3 is correct. When you changed ^(-1/3) to ^(1/3), you should bring up the denominator to the numerator; 1 / n^(-1/3) = n^(1/3).
would it be possible that someone shows me the procedure to solve it. I still have the same result...!
As the denominator is to the power -1/3 you need to root it as you have but as it is a negative power it moves up. (5a^3 b^2)(2a^-3 b)=10b^3 You were very close.
behold, my Latex skills! For I'm useless elsewhere! $\displaystyle \frac{(25a^6 b^4)^\frac{1}{2}}{(8a^{-9} b^3)^{-\frac{1}{3}}} = (25a^6 b^4)^\frac{1}{2}(8a^{-9} b^3)^{\frac{1}{3}} = (5a^3 b^2)(2a^{-3} b) = 10b^3$ ain't it pretty? :3
I must get around to learning latex. A communication essential on this forum I think.
Uppss!
Last edited by Alienis Back; Oct 2nd 2009 at 12:08 PM.
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