How do you express $\frac{1}{x^{3} - 6 x^{2} + 9 x}$ $1/(x^3-6x^2+9x)$ in partial fractions?

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How do you express $\frac{1}{x^{3} - 6 x^{2} + 9 x}$ $1/(x^3-6x^2+9x)$ in partial fractions?

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Answer 1 · 2017-01-07T10:58:27

The answer is $= \frac{\frac{1}{9}}{x} + \frac{\frac{1}{3}}{{(x - 3)}^{2}} + \frac{- \frac{1}{9}}{x - 3}$ $=(1/9)/x+(1/3)/(x-3)^2+(-1/9)/(x-3)$

Explanation:

Let's factorise the denominator

$x^{3} - 6 x^{2} + 9 x = x (x^{2} - 6 x + 9) = x {(x - 3)}^{2}$ $x^3-6x^2+9x=x(x^2-6x+9)= x(x-3)^2$

Therefore,

$\frac{1}{x^{3} - 6 x^{2} + 9 x} = \frac{1}{x {(x - 3)}^{2}}$ $1/(x^3-6x^2+9x)=1/(x(x-3)^2)$

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

$= \frac{A {(x - 3)}^{2} + B x + C x (x - 3)}{x {(x - 3)}^{2}}$ $=(A(x-3)^2+Bx+Cx(x-3))/(x(x-3)^2)$

So,

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

Let $x = 0$ $x=0$ , $\Rightarrow$ $=>$ , $1 = 9 A$ $1=9A$

Coefficients of $x^{2}$ $x^2$ ,

$0 = A + C$ $0=A+C$ , $\Rightarrow$ $=>$ , $C = - A = - \frac{1}{9}$ $C=-A=-1/9$

Let $x = 3$ $x=3$ , $\Rightarrow$ $=>$ , $1 = 3 B$ $1=3B$

Therefore,

$\frac{1}{x^{3} - 6 x^{2} + 9 x} = \frac{\frac{1}{9}}{x} + \frac{\frac{1}{3}}{{(x - 3)}^{2}} + \frac{- \frac{1}{9}}{x - 3}$ $1/(x^3-6x^2+9x)=(1/9)/x+(1/3)/(x-3)^2+(-1/9)/(x-3)$

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How do you express $\frac{1}{x^{3} - 6 x^{2} + 9 x}$ $1/(x^3-6x^2+9x)$ in partial fractions?

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Explanation:

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