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

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Answer 1 · 2016-02-23T22:15:23

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

the factors of $(x^{3} - 1)$ $(x^3-1)$ are $(x - 1)$ $(x-1)$ and $(x^{2} + x + 1)$ $(x^2+x+1)$

Set the equations using variables A, B, C so that

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

solve for the variables so that
$A = \frac{1}{3}$ $A=1/3$
$B = - \frac{1}{3}$ $B=-1/3$
$C = - \frac{2}{3}$ $C=-2/3$

God bless you ...

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