Question 1125671
You may not have meant {{{x-1/3=2}}} , which is what I read in your posting.
 
If you meant {{{x^("-1/3")=2}}} , we can "undo" that power,
and transform/simplify that equation
by replacing expressions with equivalent ones,
until we get to the solution. 
{{{x^("-1/3")=2}}}
{{{(x^("-1/3"))^-3}}}{{{"="}}}{{{2^-3}}}
{{{x^(("-1/3"(-3)))}}}{{{"="}}}{{{1/2^3}}}
{{{x^"3/3"=1/8}}}
{{{x^1=1/8}}}
{{{highlight(x=1/8)}}}
I see that as the easiest way.
Another way:
{{{x^("-1/3")}}}{{{"="}}}{{{1/x^"1/3"}}}{{{"="}}}{{{1/root(3,x)}}}{{{"="}}}{{{root(3,1/x)}}}
The equation could be written as
{{{1/root(3,x)}}}{{{"="}}}{{{2}}} , and solved with
{{{1}}}{{{"="}}}{{{2root(3,x)}}}
{{{1/2}}}{{{"="}}}{{{root(3,x)}}}
{{{(1/2)^3}}}{{{"="}}}{{{(root(3,x))^3}}}
{{{1/8=x}}} ,
or something like that.