Question 26578
you do not say whether your powers apply to all the terms or just the second in each pair?


{{{(x)^(-2)}}} is 1/(x)^2


so, {{{(b-a)^-1)}}} is 1/(b-a)
Similarly {{{(a-b)^-1)}}} is 1/(a-b)


so {{{(b-a)^(-1)/(a-b)^(-1)}}} becomes {{{(1/(b-a))/(1/(a-b))}}} which is 
{{{(a-b)/(b-a)}}}


jon.