Question 1127248

{{{4xz^(-1/3)}}} if you have only {{{z}}} to the power of {{{(-1/3)}}}, then

{{{4x(1/z^(1/3))}}} 

{{{4x/z^(1/3)}}} 

but, if you have only {{{4xz}}} to the power of {{{(-1/3)}}}, then

{{{(4xz)^(-1/3)}}} 

{{{1/(4xz)^(1/3)}}}