Chapter 11.5, Problem 21AYU

In problems 13-46, find the partial fraction decomposition of each rational expression. $\frac{1}{x^3-8}$

To determine

To find: The partial fraction decomposition of the given rational function.

Answer to Problem 21AYU

Solution:

$\frac{1}{12\left(x - 2\right)} - \frac{\left(x + 4\right)}{12\left(x^2 + 2x + 4\right)}$

Explanation of Solution

Given:

$\frac{1}{x^3 - 8}$

Formula used:

a. If $Q(x) = (x - a_1)\dots \dots .(x - a_n)$ then partial fraction decomposition of $\text{P}\left(\text{x}\right)\text{/Q}\left(\text{x}\right)$ takes the form:

$\frac{P\left(x\right)}{Q\left(x\right)} = \frac{A_1}{\left(x - a_1\right)} + \frac{A_2}{\left(x - a_2\right)} + \cdots \frac{A_n}{\left(x - a_n\right)}$

where $A_1, A_2\mathrm{...}{A}_n$ are to be determined.

b. If $Q\left(x\right)$ contains a non-repeated irreducible quadratic factor of the form $a{x}^2 + bx + c$, then, in the partial fraction decomposition of $P\left(x\right)\text{/}Q\left(x\right)$, takes the form:

$\frac{Ax + B}{a{x}^2 + bx + c}$

where the numbers $\text{A}$ and $\text{B}$ are to be determined.

Calculation:

The denominator can be factored as

$x^3 - 8 = \left(x - 2\right)\left(x^2 + 2x + 4\right)$

The given partial fraction decomposition takes the form

$\frac{1}{x^3 - 8} = \frac{A}{x - 2} + \frac{Bx + c}{\left(x^2 + 2x + 4\right)}$   -----Eq (1)

$\begin{array}{l}1 = A\left(x^2 + 2x + 4\right) + \left(Bx + C\right)\left(x - 2\right)\\ = A{x}^2 + 2Ax + 4A - 2Bx + B{x}^2 + Cx - 2C\end{array}$

Equating the coefficients of $x^2$, $x$ and constant we get

$\begin{array}{c}A + B = 0\\ 2A - 2B + C = 0\\ 4A - 2C = 1\end{array}$

Solving for $A, B, C$

$\begin{array}{l}A = 1/12\\ B = - 1/12\\ C = - 1/3\end{array}$

From Eq (1), the partial fraction decomposition of

$\frac{1}{x^3 - 8}$ is $\frac{1}{12\left(x - 2\right)} - \frac{\left(x + 4\right)}{12\left(x^2 + 2x + 4\right)}$.