Question 156848
{{{(x^-1-x^-2)^-1=(1/x^1-1/x^2)^-1=(x/x^2-1/x^2)^-1=((x-1)/(x^2))^-1=(x^2)/(x-1)}}}



So {{{(x^(-1)-x^(-2))^(-1)=(x^2)/(x-1)}}} where {{{x<>0}}} or {{{x<>1}}}