Q.Resolve $\frac{x^3}{(2x-1)(x-1)^2}$ into partial fractions.

The numerator's degree equals the denominator's degree, so first divide out a constant, then resolve the proper fraction that remains.

Denominator: $(2x-1)(x-1)^2 = 2x^3-5x^2+4x-1$. Since $\deg(\text{numerator})=\deg(\text{denominator})=3$, divide first:

$\frac{x^3}{2x^3-5x^2+4x-1} = \frac12 + \frac{\tfrac52x^2-2x+\tfrac12}{(2x-1)(x-1)^2}$

(the remainder comes from $x^3-\tfrac12(2x^3-5x^2+4x-1)=\tfrac52x^2-2x+\tfrac12$).

Now resolve the proper part:

$\frac{\tfrac52x^2-2x+\tfrac12}{(2x-1)(x-1)^2} = \frac{A}{2x-1}+\frac{B}{x-1}+\frac{C}{(x-1)^2}.$

Clearing denominators:

$\tfrac52x^2-2x+\tfrac12 = A(x-1)^2+B(2x-1)(x-1)+C(2x-1).$

At $x=1$: the $A$ and $B$ terms vanish: $\tfrac52-2+\tfrac12=1=C(1)\Rightarrow C=1$.

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