# How do you express (13x + 2)/( x^3 -1)  in partial fractions?

Mar 24, 2016

Partial fractions are $\frac{5}{\left(x - 1\right)} + \frac{- 5 x + 3}{{x}^{2} + x + 1}$

#### Explanation:

In $\frac{13 x + 2}{{x}^{3} - 1}$, we should first factorize ${x}^{3} - 1$, using identity $\left({a}^{3} + {b}^{3}\right) = \left(a + b\right) \left({a}^{2} - a b + {b}^{2}\right)$.

Hence factors are given by $\left({x}^{3} - 1\right) = \left(x - 1\right) \left({x}^{2} - x + 1\right)$.

Partial fractions of $\frac{13 x + 2}{{x}^{3} - 1}$, will be

$\frac{13 x + 2}{{x}^{3} - 1} \Leftrightarrow \frac{A}{x - 1} + \frac{B x + C}{{x}^{2} + x + 1}$and simplifying RHS it becomes

$\frac{13 x + 2}{{x}^{3} - 1} \Leftrightarrow \frac{A \left({x}^{2} + x + 1\right) + \left(B x + C\right) \left(x - 1\right)}{\left(x - 1\right) \left({x}^{2} + x + 1\right)}$ or

$\frac{13 x + 2}{{x}^{3} - 1} \Leftrightarrow \frac{\left(A {x}^{2} + A x + A\right) + \left(B {x}^{2} - B x + C x - C\right)}{{x}^{3} - 1}$ or

$\frac{13 x + 2}{{x}^{3} - 1} \Leftrightarrow \frac{\left(A + B\right) {x}^{2} + \left(A - B + C\right) x + \left(A - C\right)}{{x}^{3} - 1}$

Hence, we have $A + B = 0$ - (i), $A - B + C = 13$ - (ii) and $A - C = 2$ - (iii).

From (i) we get $B = - A$, from (iii) $C = A - 2$ and putting them in (ii),

we get $A - \left(- A\right) + \left(A - 2\right) = 13$ or $3 A = 15$ or $A = 5$

Hence $B = - 5$ and $C = 3$

Hence partial fractions are $\frac{5}{\left(x - 1\right)} + \frac{- 5 x + 3}{{x}^{2} + x + 1}$