How do you express $1/[(x^3)-1]$ in partial fractions?

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Answer 1 · 2016-03-09T23:54:32

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

Sticking with Real coefficients, we can factor the denominator as:

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

Hence we are looking for a decomposition of the form:

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

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

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

So equating coefficients we get three simultaneous equations:

$A+B=0$

$A-B+C=0$

$A-C=1$

Adding all of these three equations together we find:

$3A = 1$

Hence $A=1/3$ , $B=-1/3$ and $C=-2/3$

So:

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

How do you express 1/[(x^3)-1]1/[(x^3)-1] in partial fractions?