Question

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

Answer 1

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

Explanation:

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

We proceed to find the [partial fraction decomposition.](https://socratic.org/precalculus/matrix-row-operations/partial-fraction-decomposition-linear-denominators)

`(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)`