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Your input: perform the partial fraction decomposition of $$$\frac{1}{x^{3} \left(3 x - 2\right)}$$$

The form of the partial fraction decomposition is

$$\frac{1}{x^{3} \left(3 x - 2\right)}=\frac{A}{x}+\frac{B}{x^{2}}+\frac{C}{x^{3}}+\frac{D}{3 x - 2}$$

Write the right-hand side as a single fraction:

$$\frac{1}{x^{3} \left(3 x - 2\right)}=\frac{x^{3} D + x^{2} \left(3 x - 2\right) A + x \left(3 x - 2\right) B + \left(3 x - 2\right) C}{x^{3} \left(3 x - 2\right)}$$

The denominators are equal, so we require the equality of the numerators:

$$1=x^{3} D + x^{2} \left(3 x - 2\right) A + x \left(3 x - 2\right) B + \left(3 x - 2\right) C$$

Expand the right-hand side:

$$1=3 x^{3} A + x^{3} D - 2 x^{2} A + 3 x^{2} B - 2 x B + 3 x C - 2 C$$

Collect up the like terms:

$$1=x^{3} \left(3 A + D\right) + x^{2} \left(- 2 A + 3 B\right) + x \left(- 2 B + 3 C\right) - 2 C$$

The coefficients near the like terms should be equal, so the following system is obtained:

$$\begin{cases} 3 A + D = 0\\- 2 A + 3 B = 0\\- 2 B + 3 C = 0\\- 2 C = 1 \end{cases}$$

Solving it (for steps, see system of equations calculator), we get that $$$A=- \frac{9}{8}$$$, $$$B=- \frac{3}{4}$$$, $$$C=- \frac{1}{2}$$$, $$$D=\frac{27}{8}$$$

Therefore,

$$\frac{1}{x^{3} \left(3 x - 2\right)}=\frac{- \frac{9}{8}}{x}+\frac{- \frac{3}{4}}{x^{2}}+\frac{- \frac{1}{2}}{x^{3}}+\frac{\frac{27}{8}}{3 x - 2}$$

Answer: $$$\frac{1}{x^{3} \left(3 x - 2\right)}=\frac{- \frac{9}{8}}{x}+\frac{- \frac{3}{4}}{x^{2}}+\frac{- \frac{1}{2}}{x^{3}}+\frac{\frac{27}{8}}{3 x - 2}$$$