Question 31146
I'll give it a shot
If I read it right, it's
{{{(4^(1/4)*s^(1/3)*t^(-1/2)) / (4*s^2*t^(-3))^(-1/4)}}}
everything on the bottom can be brought up to the top just
by changing ther sign of the exponent -1/4 to +1/4
that's because {{{1 / ((4*s^2*t^(-3))^(-1/4)) = (4*s^2*t^(-3))^(1/4)}}}
so now I have
{{{(4^(1/4)*s^(1/3)*t^(-1/2)) * (4*s^2*t^(-3))^(1/4)}}}
{{{(4^(1/4)*s^(1/3)*t^(-1/2)) * 4^(1/4)*(s^2)^(1/4)*(t^(-3))^(1/4)}}}
This is because {{{(a^b *c^d)^f = a^(b*f)*c^(d*f)}}}
now I collect and combine terms
{{{4^(1/4)*4^(1/4)*s^(1/3)* (s^2)^(1/4)*t^(-1/2)*(t^(-3))^(1/4)}}}
now I use {{{a^b*a^c = a^(b+c)}}}
{{{4^(1/2)*s^(1/3)* s^(1/2)*t^(-1/2)* t^(-3/4)}}}
{{{4^(1/2)*s^(5/6)*t^(-5/4)}}} this is the solution
This actually checks out OK because
if {{{a/b = c}}} then {{{b*c = a}}}
{{{(4*s^2*t^(-3))^(-1/4)* (4^(1/2)*s^(5/6)*t^(-5/4))}}} 
should equal {{{4^(1/4)*s^(1/3)*t^(-1/2))}}}
and it actually does
{{{4^(-1/4)*s^(-1/2)*t^(3/4)*4^(1/2)*s^(5/6)*t^(-5/4)}}}=
{{{4^(-1/4)*4^(1/2)*s^(-1/2)*s^(5/6)*t^(3/4)*t^(-5/4)}}}=
{{{4^(1/4)*s^(1/3)*t^(-1/2))}}}
and that's the numerator 
hope you follow this. . . it's a good problem