Question 209264
(x-^3y^3)/(3y^-3)^2

Is this problem actually  {{{(x^-3y^3)/(3y^-3)^2}}}? I'm assuming it is.

{{{(x^-3y^3)/(3y^-3)^2}}} ----> {{{((1/x^3)*(y^3))/(9y^-6)}}} ----> {{{((1/x^3)*(y^3))/(9*(1/y^6))}}} ----> {{{((1/x^3)*(y^3))/(9/y^6))}}} ----> {{{(1/x^3)*(y^3)*(y^6/9)}}} 

Therefore, {{{(x^-3y^3)/(3y^-3)^2}}} = {{{highlight_green((y^9)/(9x^3))}}}


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Check
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Original expression
Let x = 2, and y = 3		

Then {{{(x^-3y^3)/(3y^-3)^2}}} = {{{(2^-3)*(3^3)/((3^2)*(3^-3)^2)}}} ------> {{{((1/2^3)*(3^3))/((3^2)*(3^-6))}}} -----> {{{((1/2^3)*(3^3))/((3^2)*(1/3^6))}}} -----> {{{((1/2^3)*(3^3))/((3^2)/3^6)}}} -----> {{{(1/2^3)*(3^3)*((3^6)/3^2)}}} -----> {{{(3^9)/((2^3)*(3^2))}}} -----> {{{(3^7cross(3^9))/((2^3)*(cross(3^2)))}}} ------> {{{(3^7)/(2^3)}}}


Answer
Let x = 2, and y = 3

{{{(y^9)/(9x^3)}}} = {{{(3^9)/(9*(2^3))}}} = {{{(3^9)/((3^2)*(2^3))}}} = {{{(3^7cross(3^9))/((cross(3^2))*(2^3))}}} = {{{(3^7)/(2^3)}}}