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

Dec 22, 2016

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

#### Explanation:

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

$\textcolor{w h i t e}{\frac{{y}^{2} + 1}{{y}^{3} - 1}} = \frac{A}{y - 1} + \frac{B y + C}{{y}^{2} + y + 1}$

$\textcolor{w h i t e}{\frac{{y}^{2} + 1}{{y}^{3} - 1}} = \frac{A \left({y}^{2} + y + 1\right) + \left(B y + C\right) \left(y - 1\right)}{{y}^{3} - 1}$

$\textcolor{w h i t e}{\frac{{y}^{2} + 1}{{y}^{3} - 1}} = \frac{\left(A + B\right) {y}^{2} + \left(A - B + C\right) y + \left(A - C\right)}{{y}^{3} - 1}$

Equating coefficients we get three simultaneous linear equations:

$A + B = 1$

$A - B + C = 0$

$A - C = 1$

Adding all three equations, we find:

$3 A = 2$

So $A = \frac{2}{3}$

Then from the first equation:

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

From the third equation:

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

So:

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