# How do you write the partial fraction decomposition of the rational expression (x^3-x^2+1) / (x^4-x^3)?

Dec 24, 2015

$\frac{1}{x - 1} - \frac{1}{x} ^ 2 - \frac{1}{x} ^ 3$

#### Explanation:

Factor the denominator.

$\frac{{x}^{3} - {x}^{2} + 1}{{x}^{3} \left(x - 1\right)} = \frac{A}{x} + \frac{B}{x} ^ 2 + \frac{C}{x} ^ 3 + \frac{D}{x - 1}$

Find a common denominator.

${x}^{3} - {x}^{2} + 1 = A {x}^{2} \left(x - 1\right) + B x \left(x - 1\right) + C \left(x - 1\right) + D {x}^{3}$

${x}^{3} - {x}^{2} + 1 = A {x}^{3} - A {x}^{2} + B {x}^{2} - B x + C x - C + D {x}^{3}$

${x}^{3} - {x}^{2} + 1 = {x}^{3} \left(A + D\right) + {x}^{2} \left(- A + B\right) + x \left(- B + C\right) + 1 \left(- C\right)$

Use this to write the system:

$\left\{\begin{matrix}A + D = 1 \\ - A + B = - 1 \\ - B + C = 0 \\ - C = 1\end{matrix}\right.$

Solve:

$\left\{\begin{matrix}A = 0 \\ B = - 1 \\ C = - 1 \\ D = 1\end{matrix}\right.$

Put this back in to the original expression:

$\frac{{x}^{3} - {x}^{2} + 1}{{x}^{4} - {x}^{3}} = \frac{1}{x - 1} - \frac{1}{x} ^ 2 - \frac{1}{x} ^ 3$