algebra - radicals and rational exponent problem

**mathmathmathmathmathmathm** · July 16th 2011, 08:16 AM

$\left( \frac{a^{-3/m} \cdot b^{6/n}}{a^{-6/m} \cdot b^{9/n}}\right)^{-1/3}$

thats the orginal problem, heres my work

$\left( \frac{a^{6/m} \cdot b^{6/n}}{a^{3/m} \cdot b^{9/n}}\right)^{-1/3}$

$\left( {a^{3/m} \cdot b^{-3/n}}\right)^{-1/3}$

ok iam having extreme difficulty w/ latex past this point so ill write out what i did from this step, so i flipped them again w/ the positive b^3/n on top and a^3/m on bottom, then applied the positive 1/3 to each and my wrong answer is b^n over a^m

the answer from a algebra problem solver i have is 1/a^1/m b^5/n

i have no idea where i went wrong, sorry i couldnt finish up in latex

**mathmathmathmathmathmathm** · July 16th 2011, 08:17 AM

.....................

**Plato** · July 16th 2011, 08:38 AM

Originally Posted by mathmathmathmathmathmathm

a^(6/m) * b^(9/n)^(-1/3)[/tex]

$\frac{a^(-3/m) * b^(6/n)}{a^(6/m) * b^(9/n)^(-1/3}$

If there is more that one character in an exponent you must use {}.
[tex]\left( \frac{a^{-3/m} \cdot b^{6/n}}{a^{6/m} \cdot b^{9/n}}\right)^{-1/3} [/tex]
$\left( \frac{a^{-3/m} \cdot b^{6/n}}{a^{6/m} \cdot b^{9/n}}\right)^{-1/3}$

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