Write the partial fraction decomposition of each rational expression.

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

The rational expression is,

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

Factoring the rational expression:

$\frac{1}{x^3-8}=\frac{1}{(x-2)(x^2+2x+4)}$.

Decomposition of the partial fractions,

role="math" localid="1647460674458" $\frac{1}{(x-2)(x^2+2x+4)}=\frac{A}{x-2}+\frac{Bx+C}{x^2+2x+4}.........\left(1\right)$.

Multiplying both sides by $(x-2)(x^2+2x+4)$,

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

Equating the coefficients of the like powers of $x$, we get,

$$\begin{gathered} A+B=0...........\left(3\right) \\ 2A-2B+C=0.........\left(4\right) \\ 4A-2C=1 \\ \Rightarrow 2A-C=\frac{1}{2}...........\left(5\right) \end{gathered}$$

$$\begin{gathered} A+B=0 \\ B=-A........\left(6\right) \end{gathered}$$

$$\begin{gathered} 2A-C=\frac{1}{2} \\ \Rightarrow C=2A-\frac{1}{2}......\left(7\right) \end{gathered}$$

Inputting the values in the equation$2A-2B+C=0$,

$$\begin{gathered} 2A-2(-A)+(2A-\frac{1}{2})=0 \\ 2A+2A+2A=\frac{1}{2} \\ 6A=\frac{1}{2} \\ A=\frac{1}{12} \end{gathered}$$

The other values are,

$$\begin{gathered} B=\frac{-1}{12} \\ C=\frac{1}{3} \end{gathered}$$

The partial fraction decomposition is,

role="math" localid="1647461553968" $\frac{1}{x^3-8}=\frac{{\displaystyle \frac{1}{12}}}{(x-2)}+\frac{{\displaystyle \frac{-1}{12}}x+{\displaystyle \frac{1}{3}}}{x^2+2x+4}$

$=\frac{1}{12(x-2)}+\frac{-1(x+4)}{12(x^2+2x+4)}$.

Adding the rational expressions,

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

$$\begin{gathered} =\frac{x^2+2x+4-x^2-4x+2x+8}{12(x^3-8)} \\ =\frac{12}{12(x^3-8)} \\ =\frac{1}{x^3-8} \end{gathered}$$

The solution is true.