Question 187307
{{{( (x*y^5) / (x^-4*y^2) )^-2 = 1 / ( (x*y^5) / (x^-4*y^2) )^2 }}}
This is  because {{{a^(-b) = 1/(a^b)}}}
{{{ 1 / ( (x*y^5) / (x^-4*y^2) )^2 =  1 / ( (x*y^5)^2 / (x^-4*y^2)^2 ) }}}
This is becaue {{{(a/b)^c = (a^c) / (b^c)}}}
{{{ 1 / ( (x*y^5)^2 / (x^-4*y^2)^2 ) = (x^-4*y^2)^2 / (x*y^5)^2 }}}
This is because {{{1/(a/b) = b/a}}}
{{{(x^-4*y^2)^2 / (x*y^5)^2 = ((y^2 / x^4)^2) / (x*y^5)^2 }}}
{{{((y^2 / x^4)^2) / (x*y^5)^2 = ((y^2 / x^4) / (x*y^5))^2}}}
This is because {{{a^c / b^c = (a/b)^c}}}
{{{((y^2 / x^4) / (x*y^5))^2 = (y^2) / (x^4*x*y^5))^2}}}
This is because {{{(a/b)/c = a/(b*c)}}}
Now divide top & bottom by {{{y^2}}}
{{{ (y^2 / (x^4*x*y^5))^2 = 1/(x^5*y^3)^2 }}} answer
-----------------------------------------------------
There is a quick  & easy way to test answer
I'll say {{{x = 3}}} and {{{y = 7}}} (I could use any numbers)
{{{( (x*y^5) / (x^-4*y^2) )^-2}}}
{{{( (3*7^5) / (3^-4*7^2) )^-2 }}}
{{{( 50421 / .604938 )^-2 }}}
{{{143.9459}}}
and
{{{1/(x^5*y^3)^2}}}
{{{1/(243*343)^2}}}
{{{143.9459}}}
OK