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

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 ${x}^{2} + x = x \left(x + 1\right)$.
Third denominator is ${x}^{3} - {x}^{2} = x \left({x}^{2} - x\right) = x \cdot x \left(x - 1\right)$.
The least common multiple is $x \cdot x \cdot \left(x + 1\right) \left(x - 1\right) = {x}^{4} - {x}^{2}$.

For first fraction to have the least common multiple as the denominator we multiply both the numerator and the denominator by ${x}^{3} - x$.
$\frac{1 \cdot \left({x}^{3} - x\right)}{x \cdot \left({x}^{3} - x\right)} = \frac{{x}^{3} - x}{{x}^{4} - {x}^{2}}$.
For second fraction we multiply both the numerator and the denominator by ${x}^{2} - x$.
$\frac{2 \left({x}^{2} - x\right)}{\left({x}^{2} + x\right) \left({x}^{2} - x\right)} = \frac{2 {x}^{2} - 2 x}{{x}^{4} - {x}^{2}}$.
For third fraction we multiply both the numerator and the denominator by $x + 1$.
$\frac{3 \left(x + 1\right)}{\left({x}^{3} - {x}^{2}\right) \left(x + 1\right)} = \frac{3 x + 3}{{x}^{4} - {x}^{2}}$.

Putting all these together:
$\frac{{x}^{3} - x}{{x}^{4} - {x}^{2}} - \frac{2 {x}^{2} - 2 x}{{x}^{4} - {x}^{2}} + \frac{3 x + 3}{{x}^{4} - {x}^{2}} = \frac{{x}^{3} - 2 {x}^{2} + 4 x + 3}{{x}^{4} - {x}^{2}}$