Answer
$\dfrac{3x-5}{(x-1)(x^2+x+1)}=-\dfrac{2}{3(x-1)}+\dfrac{2x+13}{3(x^2+x+1)}$
Work Step by Step
We are given the fraction:
$\dfrac{3x-5}{x^3-1}$
Factor the denominator:
$\dfrac{3x-5}{x^3-1}=\dfrac{3x-5}{(x-1)(x^2+x+1)}$
We can write the partial fraction decomposition:
$\dfrac{3x-5}{(x-1)(x^2+x+1)}=\dfrac{A}{x-1}+\dfrac{Bx+C}{x^2+x+1}$
Multiply all terms by the least common denominator $(x-1)(x^2+x+1)$:
$(x-1)(x^2+x+1)\cdot\dfrac{3x-5}{(x-1)(x^2+x+1)}=(x-1)(x^2+x+1)\cdot\dfrac{A}{x-1}+(x-1)(x^2+x+1)\cdot\dfrac{Bx+C}{x^2+x+1}$
$3x-5=A(x^2+x+1)+(Bx+C)(x-1)$
$3x-5=Ax^2+Ax+A+Bx^2-Bx+Cx-C$
$3x-5=(A+B)x^2+(A-B+C)x+(A-C)$
Identify the coefficients and build the system of equations:
$\begin{cases}
A+B=0\\
A-B+C=3\\
A-C=-5
\end{cases}$
Solve the system: Add Equation 2 to Equation 3:
$\begin{cases}
A+B=0\\
A-B+C+A-C=3+(-5)
\end{cases}$
$\begin{cases}
A+B=0\\
2A-B=-2
\end{cases}$
$A+B+2A-B=0-2$
$3A=-2$
$A=-\dfrac{2}{3}$
$-\dfrac{2}{3}+B=0$
$B=\dfrac{2}{3}$
$A-C=-5$
$-\dfrac{2}{3}-C=-5$
$C=5-\dfrac{2}{3}$
$C=\dfrac{13}{3}$
The partial fraction decomposition is:
$\dfrac{3x-5}{(x-1)(x^2+x+1)}=\dfrac{-\dfrac{2}{3}}{x-1}+\dfrac{\dfrac{2}{3}x+\dfrac{13}{3}}{x^2+x+1}$
$\dfrac{3x-5}{(x-1)(x^2+x+1)}=-\dfrac{2}{3(x-1)}+\dfrac{2x+13}{3(x^2+x+1)}$
