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$ \left(\dfrac{a^{2}b^{-{\color{red}3}}}{x^{-{\color{red}3}}y^{2}}\right)^{3}\left(\dfrac{x^{-2}b^{-1}}{a^{\frac{{\color{red}3}}{{\color{red}4}}}y^{\frac{1}{3}}}\right) $

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Simplify to what? You just want to combine all the powers of the same term? Then you will have $a^{6-\frac{3}{4}}b^{-10}x^{7}y^{-6-\frac{1}{3}}$ –  Dennis Gulko Sep 9 '12 at 16:15
Welcome to Math SE. While we do enjoy answering questions here, we do respond better if you ask nicely. Also it really helps if you show us what you have already done, as it shows you have spent some effort on it before giving up and asking us to do it for you. –  Arthur Sep 9 '12 at 16:16
Right and eliminate the negative exponents. –  Brandt Sep 9 '12 at 16:18

2 Answers 2

Use the laws of exponents: You can bring the cube on the first fraction inside using $(c^d)^e=c^{de}$. Then pull all the terms in denominators upstairs using $\frac 1a=a^{-1}$. Then combine the powers of the same variable using $c^dc^e=c^{(d+e)}$

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What you want is probably $$a^{5 \frac{1}{4}} b^{-10} x^{7} y^{-6 \frac{1}{3}}$$ You just have to add up the powers...

Looks very much like high school homework.

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