(2x ^{-3} yz ^{-6}) (2x)^{-5} (I'm little lost on how to get y/16^8 z^6) ?.? only at this part (-32x^-5)/2x^-3yz^-6

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(2x ^{-3} yz ^{-6}) (2x)^{-5} (I'm little lost on how to get y/16^8 z^6) ?.? only at this part (-32x^-5)/2x^-3yz^-6

Mathematics
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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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|dw:1327351420225:dw|
Thanks Pross
:)

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(2^{-1}x ^{-2}y ^{-1})^{-2}(2x ^{-3}y)^{-2}(16x ^{-3}y ^{3})^{0}/ (2x ^{-3}y ^{-5)^{2}}
first multiply each exponent in the first large parentheses times the -2. Then in the second large parentheses the zero exponent will make the whole term equal to 1.
that should get you started.
I starting to understand a little, now I'm little confuse on this part (4x^4y^2) (2x^6 y ^-2) (xy)/ 4x^-6 y^-10. ?.?
|dw:1327353996471:dw| I am not sure that I have all of your parentheses correctly written out...take a look and see...

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