How do you express #1/[(x^3)-1]# in partial fractions?
1 Answer
Mar 9, 2016
#1/(x^3-1) = 1/(3(x-1))-(x+2)/(3(x^2+x+1))#
Explanation:
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
So:
#1/(x^3-1) = 1/(3(x-1))-(x+2)/(3(x^2+x+1))#