How do you simplify #(1)/(x) - (2)/(x^2 + x) + (3)/(x^3 - x^2)#?

1 Answer
Oct 9, 2015

The key to this is to make the denominator the same for all 3 fractions. For this, you have to find the least common multiple of the 3 denominators.
First denominator is just #x#.
Second denominator is #x^2+x=x(x+1)#.
Third denominator is #x^3-x^2=x(x^2-x)=x*x(x-1)#.
The least common multiple is #x*x*(x+1)(x-1)=x^4-x^2#.

For first fraction to have the least common multiple as the denominator we multiply both the numerator and the denominator by #x^3-x#.
#(1*(x^3-x))/(x*(x^3-x))=(x^3-x)/(x^4-x^2)#.
For second fraction we multiply both the numerator and the denominator by #x^2-x#.
#(2(x^2-x))/((x^2+x)(x^2-x))=(2x^2-2x)/(x^4-x^2)#.
For third fraction we multiply both the numerator and the denominator by #x+1#.
#(3(x+1))/((x^3-x^2)(x+1))=(3x+3)/(x^4-x^2)#.

Putting all these together:
#(x^3-x)/(x^4-x^2)-(2x^2-2x)/(x^4-x^2)+(3x+3)/(x^4-x^2)=(x^3-2x^2+4x+3)/(x^4-x^2)#