# How do you write the partial fraction decomposition of the rational expression  (y^2 + 1) / (y^3 - 1)?

Feb 28, 2017

The answer is $= \frac{\frac{2}{3}}{y - 1} + \frac{\frac{1}{3} y - \frac{1}{3}}{{y}^{2} + y + 1}$

#### Explanation:

Let's factorise the denominator

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

Therefore,

$\frac{{y}^{2} + 1}{{y}^{3} - 1} = \frac{{y}^{2} + 1}{\left(y - 1\right) \left({y}^{2} + y + 1\right)}$

Let's perform the decomposition into partial fractions

$\frac{{y}^{2} + 1}{\left(y - 1\right) \left({y}^{2} + y + 1\right)} = \frac{A}{y - 1} + \frac{B y + C}{{y}^{2} + y + 1}$

$= \frac{A \left({y}^{2} + y + 1\right) + \left(B y + C\right) \left(y - 1\right)}{\left(y - 1\right) \left({y}^{2} + y + 1\right)}$

The denominators are the same, we can compare the numerators

${y}^{2} + 1 = A \left({y}^{2} + y + 1\right) + \left(B y + C\right) \left(y - 1\right)$

Let $y = 1$, $\implies$, $2 = 3 A$, $\implies$, $A = \frac{2}{3}$

Coefficients of ${y}^{2}$

$1 = A + B$, $\implies$, $B = 1 - A = 1 - \frac{2}{3} = \frac{1}{3}$

Also,

$1 = A - C$, $\implies$, $C = A - 1 = \frac{2}{3} - 1 = - \frac{1}{3}$

Therefore,

$\frac{{y}^{2} + 1}{\left(y - 1\right) \left({y}^{2} + y + 1\right)} = \frac{\frac{2}{3}}{y - 1} + \frac{\frac{1}{3} y - \frac{1}{3}}{{y}^{2} + y + 1}$