Question 103489
([2a^2b]/[3x^2y])^3  times  ([-6xy^2]/[a^3b])^2
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=([8a^6b^3] / [27x^6y^3]) * ([36x^2y^4] / [a^6b^2])

Get the ab's together and the xy's together:

=([8a^6b^3]  / [a^6b^2]) * ([36x^2y^4] / [27x^6y^3]) 

Cancel where you can to get:

= (8b)*(4y/3x^4)

= [32by/3x^4]

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Cheers,
Stan H.