How do you write the partial fraction decomposition of the rational expression $\frac{1}{(x^{4}) - 6 (x^{3})}$ $1/((x^4)-6(x^3))$ ?

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Answer 1 · 2015-12-14T16:16:34

$\frac{1}{216 (x - 6)} - \frac{1}{216 x} - \frac{1}{36 x^{2}} - \frac{1}{6 x^{3}}$ $1/(216(x-6))-1/(216x)-1/(36x^2)-1/(6x^3)$

Factor the denominator.

$\frac{1}{x^{3} (x - 6)} = \frac{A}{x} + \frac{B}{x^{2}} + \frac{C}{x^{3}} + \frac{D}{x - 6}$ $1/(x^3(x-6))=A/x+B/x^2+C/x^3+D/(x-6)$

$1 = A (x^{2}) (x - 6) + B (x) (x - 6) + C (x - 6) + D (x^{3})$ $1=A(x^2)(x-6)+B(x)(x-6)+C(x-6)+D(x^3)$

$1 = A x^{3} - 6 A x^{2} + B x^{2} - 6 B x + C x - 6 C + D x^{3}$ $1=Ax^3-6Ax^2+Bx^2-6Bx+Cx-6C+Dx^3$

$1 = x^{3} (A + D) + x^{2} (- 6 A + B) + x (- 6 B + C) + 1 (- 6 C)$ $1=x^3(A+D)+x^2(-6A+B)+x(-6B+C)+1(-6C)$

Use this to write the following system:

${(A+D=0),(-6A+B=0),(-6B+C=0),(-6C=1):}$

Solve to see that:

${(A=-1/216),(B=-1/36),(C=-1/6),(D=1/216):}$

The fraction decomposes into:

$1/(216(x-6))-1/(216x)-1/(36x^2)-1/(6x^3)$

How do you write the partial fraction decomposition of the rational expression 1/((x^4)-6(x^3))1(x4)−6(x3) 1/((x^4)-6(x^3))?