Answer
The simplified partial fraction expansion is $\frac{\left( -2 \right)}{3\left( x-1 \right)}+\frac{2x+13}{3\left( {{x}^{2}}+x+1 \right)}$.
Work Step by Step
The provided partial fraction expression is as follows:
$\frac{3x-5}{{{x}^{3}}-1}$
Simplify this as given below:
$\begin{align}
& \frac{3x-5}{{{x}^{3}}-1}=\frac{3x-5}{{{x}^{3}}-{{1}^{3}}} \\
& =\frac{3x-5}{\left( x-1 \right)\left( {{x}^{2}}+x+1 \right)}
\end{align}$
Step 1:
Set up the partial fraction expansion with unknown constant coefficients and then write a constant coefficient over each of the two distinct algebraic linear factors in the denominator of the expression.
$\frac{3x-5}{\left( x-1 \right)\left( {{x}^{2}}+x+1 \right)}=\frac{A}{x-1}+\frac{Bx+C}{{{x}^{2}}+x+1}$
Step 2:
Multiply both sides of the equation by the expression $\left( x-1 \right)\left( {{x}^{2}}+x+1 \right)$ and consider the least common denominator.
$\left( x-1 \right)\left( {{x}^{2}}+x+1 \right)\times \left( \frac{3x-5}{\left( x-1 \right)\left( {{x}^{2}}+x+1 \right)} \right)=\left( x-1 \right)\left( {{x}^{2}}+x+1 \right)\times \left( \frac{A}{x-1}+\frac{Bx+C}{{{x}^{2}}+x+1} \right)$
Then, multiply and simplify as follows:
$\begin{align}
& 3x-5=\left( {{x}^{2}}+x+1 \right)A+\left( Bx+C \right)\left( x-1 \right) \\
& 3x-5=\left( {{x}^{2}}+x+1 \right)A+\left( Bx+C \right)\left( x-1 \right) \\
& =A\left( {{x}^{2}}+x+1 \right)+B\left( {{x}^{2}}-x \right)+C\left( x-1 \right) \\
& ={{x}^{2}}\left( A+B \right)+x\left( A-B+C \right)+A-C
\end{align}$
Step 3:
And equating the coefficients of like powers of $ x $ and of the constant terms. Then, we get the system of linear equations with the unknown values of $ A,B $ and $ C $.
$ A+B=0$ (I)
$ A-B+C=3$ (II)
$ A-C=-5$ (III)
Step 5:
Now solve the system for $ A $, $ B $, and $ C $,
Eliminate $ C $ by adding equations (II) and (III) as given below:
$\left( A-B+C \right)+A-C=3-5$
$2A-B=-2$ (IV)
Eliminate $ B $ by adding equations (I) and (IV) as given below:
$\begin{align}
& \left( A+B \right)+2A-B=0-2 \\
& 3A=-2 \\
& A=-\frac{2}{3}
\end{align}$
Substituting the values of $ A $ in equation (I) and simplify as given below:
$\begin{align}
& A+B=0 \\
& -\frac{2}{3}+B=0 \\
& B=\frac{2}{3}
\end{align}$
Similarly, put the values of $ A $ and $ B $ in equation (II), and find the value of $ C $ as given below:
$\begin{align}
& A-B+C=3 \\
& -\frac{2}{3}-\frac{2}{3}+C=3 \\
& C=3+\frac{4}{3} \\
& C=\frac{13}{3}
\end{align}$
Step 5:
By replacing the values of $ A $, $ B $, and $ C $, write the partial function expression as given below:
$\begin{align}
& \frac{3x-5}{\left( x-1 \right)\left( {{x}^{2}}+x+1 \right)}=\frac{A}{x-1}+\frac{Bx+C}{{{x}^{2}}+x+1} \\
& =\frac{-\frac{2}{3}}{x-1}+\frac{\frac{2}{3}x+\frac{13}{3}}{{{x}^{2}}+x+1} \\
& =-\frac{2}{3\left( x-1 \right)}+\frac{2x+13}{3\left( {{x}^{2}}+x+1 \right)}
\end{align}$
Thus, $-\frac{2}{3\left( x-1 \right)}+\frac{2x+13}{3\left( {{x}^{2}}+x+1 \right)}$ is the required partial fraction expansion of the rational expression $\frac{3x-5}{{{x}^{3}}-1}$.
