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

Since the numerator's degree is not less than the denominator's, first divide out the polynomial part, then resolve the proper-fraction remainder into partial fractions.

Step 1 — Divide (degree of numerator $\ge$ degree of denominator).

$(x-1)(x+2)=x^2+x-2$. Divide $x^3$ by $x^2+x-2$:

$x^3 = (x^2+x-2)(x-1) + (3x-2)$ (long division).

So $\dfrac{x^3}{(x-1)(x+2)} = (x-1) + \dfrac{3x-2}{(x-1)(x+2)}$.

Step 2 — Resolve the proper fraction $\dfrac{3x-2}{(x-1)(x+2)}$.

Let $\dfrac{3x-2}{(x-1)(x+2)}=\dfrac{A}{x-1}+\dfrac{B}{x+2}$, so $3x-2=A(x+2)+B(x-1)$.

Put $x=1$: $3(1)-2=1=A(3)\implies A=\dfrac13$.

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