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))#