Question 601287
You should have (y^-5/3x^-3)^2  because (y^-1)/(y^4) = y^(-1-4) = y^(-5)



Now square both the numerator and denominator separately


Square the numerator: (y^(-5))^2 = y^(-5*2) = y^(-10)


So the new numerator is now  y^(-10). To write the exponent as a positive number, you flip the base to get 1/(y^10)


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Square the denominator: (3x^-3)^2 = 3^2*x^(-3*2) = 9x^-6


So the new denominator is now 9x^-6



Finally, flip the base that has the negative exponent to get  9/(x^6)



So combine the new numerator and denominator to get



(1/(y^10))/(9/(x^6)) = (1/(y^10))*(x^6)/9 = (x^6)/(9y^10)




Giving you the final answer of (x^6)/(9y^10)



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Answer:



So  <img src="http://latex.codecogs.com/gif.latex?\LARGE \left(\frac{24y^{-1}}{72x^{-3}y^4}\right)^2" title="x" /> completely simplifies to  <img src="http://latex.codecogs.com/gif.latex?\LARGE \frac{x^6}{9y^{10}}" title="x" />



In other words <img src="http://latex.codecogs.com/gif.latex?\LARGE \left(\frac{24y^{-1}}{72x^{-3}y^4}\right)^2 = \frac{x^6}{9y^{10}}" title="x" /> for all allowed values of x and y.