Question 1112786
It depends on what you know about {{{a}}} and {{{n}}}.
If you are given values for {{{a}}} and {{{n}}} it is easy:
 
If {{{a>=0}}} and {{{n}}} is a positive integer, {{{root(n,a^n)=a}}} .
Example #1: for {{{a=2}}} and {{{n=7}}} , {{{root(7,2^7)=root(7,128)=2}}} .
Example #2: for {{{a=3}}} and {{{n=4}}} , {{{root(6,3^4)=root(4,81)=3}}} .
 
If {{{a<0}}} and {{{n}}} is an even (positive) integer, {{{root(n,a^n)=abs(a)}}} .
Example #3: for {{{a=-5}}} and {{{n=4}}} , {{{root(4,(-5)^4)=root(4,625)=5}}} . 
 
If {{{a<0}}} and {{{n}}} is an odd (positive) integer, {{{root(n,a^n)=a}}} .
Example #3: for {{{a=-5}}} and {{{n=3}}} , {{{root(3,(-5)^3)=root(4,-125)=-5}}} .
 

 
IF YOU DO NOT HAVE ENOUGH INFORMATION: 
If {{{n}}} is an even (positive) integer, {{{root(n,a^n)=abs(a)}}} is always true.
{{{root(4,(-5)^4)=root(4,625)=5=abs(-5)}}} and {{{root(4,5^4)=root(4,625)=5=abs(-5)}}} .
 
If {{{n}}} is an odd (positive) integer, {{{root(n,a^n)=a}}} is always true.
{{{root(3,(-5)^43)=root(4,-125)=-5}}} and {{{root(3,5^3)=root(4,125)=5}}} .
 
By definition of negative exponent, {{{a^(-"2 / 3")}}}{{{"="}}}{{{1/a^"2 / 3"}}} ,
and {{{1/"-27 / 64"}}}{{{"="}}}{{{-1/"27 / 64"}}}{{{"="}}}{{{(-1)(64/27)}}}{{{"="}}}{{{-64/27}}} as you already knew.
You probably would just write as one step: {{{1/"-27 / 64"}}}{{{"="}}}{{{-64/27}}} .
 
Putting that old and new knowledge together, what you are probably expected to write is
{{{(-27/64)^"- 2 / 3"}}}{{{"="}}}{{{(-64/27)^"2 / 3"}}}{{{"="}}}{{{(root(3,-64/27))^2)}}}{{{"="}}}{{{(-4/3)^2}}}{{{"="}}}{{{(16/9)}}}