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

Dec 23, 2015

(x^3-x^2+1) / (x^4-x^3) = -1/x^3-1/x^2+1/(x-1

#### Explanation:

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

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

$= \frac{x - 1}{x \left(x - 1\right)} + \frac{1}{{x}^{3} \left(x - 1\right)}$

$= \frac{1}{x} + \frac{1}{{x}^{3} \left(x - 1\right)}$

now focus on $\frac{1}{{x}^{3} \left(x - 1\right)}$

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

Multiply both side by ${x}^{3} \left(x - 1\right)$

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

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

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

$C + D = 0$
$B - C = 0$
$A - B = 0$
$A = - 1$

Just by looking we have

$A = - 1$
$B = - 1$
$C = - 1$
$D = 1$

So

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