Question 1007160
{{{f(x)= 1/(2x^3-9)}}} 

find: {{{f^-1(x)}}}

recall {{{f(x)= y}}} 


{{{y= 1/(2x^3-9)}}}........swap {{{x}}} and {{{y}}} 


{{{x= 1/(2y^3-9)}}}........solve for {{{y}}}


{{{x(2y^3-9)= 1}}}


{{{2y^3-9= 1/x}}}


{{{2y^3= 1/x+9}}}


{{{y^3= (1/x+9)/2}}}


{{{y= root(3,(1/x+9)/2)}}}


{{{y= root(3,1/x+9)/root(3,2)}}}


so, your inverse is:  {{{f^-1(x)=root(3,1/x+9)/root(3,2)}}}