How do you express #(8x-1)/(x^3 -1)# in partial fractions?

1 Answer
George C.
Dec 8, 2016

#(8x-1)/(x^3-1) = 7/(3(x-1)) - (7x-10)/(3(x^2+x+1))#

Explanation:

Note that #x^3-1 = (x-1)(x^2+x+1)#

So:

#(8x-1)/(x^3-1) = A/(x-1) + (Bx+C)/(x^2+x+1)#

#color(white)((8x-1)/(x^3-1)) = (A(x^2+x+1) + (Bx+C)(x-1))/(x^3-1)#

#color(white)((8x-1)/(x^3-1)) = ((A+B)x^2+(A-B+C)x+(A-C))/(x^3-1)#

Equating coefficients we find:

#{ (A+B = 0), (A-B+C = 8), (A-C = -1) :}#

Adding all three equations, we find:

#3A = 7#

So:

#A=7/3#

From the first equation we can deduce:

#B = -A = -7/3#

From the third equation:

#C = A+1 = 7/3+1 = 10/3#

So:

#(8x-1)/(x^3-1) = 7/(3(x-1)) - (7x-10)/(3(x^2+x+1))#

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