How do you express $(x^3-x^2+1) / (x^4-x^3)$ in partial fractions?

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Answer 1 · 2017-05-22T17:45:37

The answer is $=-1/x^3-1/(x^2)+1/(x-1)$

The denominator is

$x^4-x^3=x^3(x-1)=x(x-1)$

Let's perform the decomposition into partial fractions

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

$=A/x^3+B/(x^2)+(C)/x+D/(x-1)$

$=(A(x-1)+Bx(x-1)+Cx^2(x-1)+Dx^3)/(x^3(x-1))$

The denominators are the same, we compare the numerators

$x^3-x^2+1=A(x-1)+Bx(x-1)+Cx^2(x-1)+Dx^3$

Let $x=0$ , $=>$ , $1=-A$ , $=>$ , $A=-1$

Let $x=1$ , $=>$ , $1=D$

Coefficients of $x^2$ ,

$-1=B-C$

Coefficients of $x^3$ ,

$1=C+D$ , $=>$ , $C=1-D=1-1=0$

$B=C-1=0-1=-1$

Therefore,

$(x^3-x^2+1)/(x^4-x^3)=-1/x^3-1/(x^2)+(0)/x+1/(x-1)$

How do you express (x^3-x^2+1) / (x^4-x^3)(x^3-x^2+1) / (x^4-x^3) in partial fractions?