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

Nov 9, 2016

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

#### Explanation:

The denominator is ${\left(x + 1\right)}^{3}$
$\therefore \frac{{x}^{2} + 2}{x + 1} ^ 3 = \frac{A}{x + 1} ^ 3 + \frac{B}{x + 1} ^ 2 + \frac{C}{x + 1}$
$= \frac{A + B \left(x + 1\right) + C {\left(x + 1\right)}^{2}}{x + 1} ^ 3$

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

let $x = - 1$$\implies$$3 = A$

If $x = 0$, $2 = A + B + C$
coefficients of ${x}^{2}$$\implies$$1 = C$
coefficients of $x$$\implies$$0 = B + 2 C$$\implies$$B = - 2$

So, $\frac{{x}^{2} + 2}{x + 1} ^ 3 = \frac{3}{x + 1} ^ 3 - \frac{2}{x + 1} ^ 2 + \frac{1}{x + 1}$