How do you write the partial fraction decomposition of the rational expression # (6x)/(x^3-8)#?

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Answer 1 · 2016-02-29T23:34:16

#(6x)/((x-2)(x^2+2x+4)) = 1/(x-2)-(x-2)/(x^2+2x+4)#

Noting that from the difference of cubes formula,
#x^3-8 = (x-2)(x^2+2x+4)#

#(6x)/((x-2)(x^2+2x+4)) = A/(x-2)+(Bx+C)/(x^2+2x+4)#

#=>6x = A(x^2+2x+4) + (Bx+C)(x-2)#

#=(A+B)x^2+(2A-2B+C)x+(4A-2C)#

Equating corresponding coefficients, we obtain the following system:

#{(A+B=0),(2A-2B+C=6),(4A-2C=0):}#

Solve this using your favorite technique to find that it resolves to

#{(A=1),(B=-1),(C=2):}#

Substituting these back in gives our final result.

#(6x)/((x-2)(x^2+2x+4)) = 1/(x-2)-(x-2)/(x^2+2x+4)#