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

The form of the partial fraction decomposition is

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

Write the right-hand side as a single fraction:

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

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

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

Expand the right-hand side:

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

Collect up the like terms:

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

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

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

Solving it (for steps, see system of equations calculator), we get that $$$A=-1$$$, $$$B=-1$$$, $$$C=1$$$

Therefore,

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

Answer: $$$\frac{1}{x^{2} \left(x - 1\right)}=\frac{-1}{x}+\frac{-1}{x^{2}}+\frac{1}{x - 1}$$$