How do you write the partial fraction decomposition of the rational expression `(8x-1)/(x^3 -1)`?

The first thing that has to be done is to factorise the denominator.
`x^3 - 1` is a difference of cubes and is factorised as follow :

`a^3 - b^3 = (a - b )(a^2 + ab + b^2 )`

here then `a = x` and `b = 1` .

So `x^3 - 1 = (x - 1 )(x^2 + x + 1 )`

Now `(x - 1)` is of degree 1 and so numerator will be of degree 0 ie. a constant. Similarly `(x^2 +x +1 )` is of degree 2 and so numerator will be of degree 1 ie of the form `ax + b`. Now the expression can be written as :

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

Multiplying through by `(x^3 - 1 )`

`8x - 1 = A(x^2 + x + 1 ) + (Bx + C )(x - 1 )" " " " color(red)(("*"))`

We now have to find the values of `A` , `B` and `C`.

Note that if we use `x = 1` then the term with A will be 0.

Substitute x = 1 in equation `color(red)(("*") )`

`7 = 3A + 0 rArr A = 7/3`

To find B and C it will be necessary to compare the coefficients on both sides of the equation `color(red)(("*"))`. Multiplying out the brackets on the right hand side to begin with.

`rArr 8x -1 = Ax^2 + Ax + A + Bx^2 + Cx - Bx - C`

This can be 'tidied up' by collecting like terms and letting `A =7/3`

`8x - 1 = 7/3 x^2 + 7/3 x + 7/3 + Bx^2 + Cx - Bx - C`

`rArr 8x - 1 = (7/3 + B)x^2 + (7/3 +C - B)x +(7/3 - C )`

Compare `x^2` terms

`0 = 7/3 + B rArr B = -7/3`

Now compare constant terms.

`- 1 = A - C = -7/3 - C rArr C= -10/3`

Finally

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