Question

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

Answer 1

`(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))`