remember to add/subtract fractions we need a common denominator.
let's write our expression in factored form
1/x - 2/(x^2 + x) + 3/(x^3 - x^2)
= 1/x - 2/[x(x+1)] + 3/[x^2(x-1)]
so the common denominator is x^2(x+1)(x-1)
first term: x(x+1)(x-1)/[x^2(x+1)(x-1)]
notice if we cancel we get 1/x
second term: -2x(x-1)/[x^2(x+1)(x-1)]
again, see what you get when you cancel
third term: 3(x+1)/[x^2(x+1)(x-1)]
so we get
[x(x+1)(x-1) - 2x(x-1) + 3(x+1)]/[x^2(x+1)(x-1)]
I will let you finish it,
(I got (x^3 - 2x^2 + 4x + 3)/[x^2(x^2 - 1)]
I bet you have a sign wrong but will do as shown.
we need a common denominator
need LCD = x^2 (x-1)(x+1) = x^2(x^2-1)
-2 [ (x)(x-1) ] /LCD
+3 [ (x+1 ] / LCD
[1 (x(x^2-1) -2 (x^2-x) +3 (x+1) ]/LCD
[ x^3-x -2x^2 +2x +3x+3 ]/LCD
(x^3 -2x^2 +4x +3)/[x^2(x^2-1)]
I could not simplify, suspct mistake but that is the idea.
Oh, I see Reiny got the same answer. If you typed it right the answer is right.
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