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

Jan 9, 2016

$\frac{\frac{7}{3}}{x - 1} + \frac{- \frac{7}{3} x - \frac{10}{3}}{{x}^{2} + x + 1}$

#### Explanation:

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 :

${a}^{3} - {b}^{3} = \left(a - b\right) \left({a}^{2} + a b + {b}^{2}\right)$

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

So ${x}^{3} - 1 = \left(x - 1\right) \left({x}^{2} + x + 1\right)$

Now $\left(x - 1\right)$ is of degree 1 and so numerator will be of degree 0 ie. a constant. Similarly $\left({x}^{2} + x + 1\right)$ is of degree 2 and so numerator will be of degree 1 ie of the form $a x + b$. Now the expression can be written as :

$\frac{8 x - 1}{{x}^{3} - 1} = \frac{A}{x - 1} + \frac{B x + C}{{x}^{2} + x + 1}$

Multiplying through by $\left({x}^{3} - 1\right)$

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 $\textcolor{red}{\left(\text{*}\right)}$

$7 = 3 A + 0 \Rightarrow A = \frac{7}{3}$

To find B and C it will be necessary to compare the coefficients on both sides of the equation $\textcolor{red}{\left(\text{*}\right)}$. Multiplying out the brackets on the right hand side to begin with.

$\Rightarrow 8 x - 1 = A {x}^{2} + A x + A + B {x}^{2} + C x - B x - C$

This can be 'tidied up' by collecting like terms and letting $A = \frac{7}{3}$

$8 x - 1 = \frac{7}{3} {x}^{2} + \frac{7}{3} x + \frac{7}{3} + B {x}^{2} + C x - B x - C$

$\Rightarrow 8 x - 1 = \left(\frac{7}{3} + B\right) {x}^{2} + \left(\frac{7}{3} + C - B\right) x + \left(\frac{7}{3} - C\right)$

Compare ${x}^{2}$ terms

$0 = \frac{7}{3} + B \Rightarrow B = - \frac{7}{3}$

Now compare constant terms.

$- 1 = A - C = - \frac{7}{3} - C \Rightarrow C = - \frac{10}{3}$

Finally

$\frac{8 x - 1}{{x}^{3} - 1} = \frac{\frac{7}{3}}{x - 1} + \frac{- \frac{7}{3} x - \frac{10}{3}}{{x}^{2} + x + 1}$