Q.Resolve $\dfrac{x^3}{(x-1)(x-2)}$ into partial fractions (reduce first, since the fraction is improper).

The numerator's degree (3) is not less than the denominator's degree (2), so this fraction is improper and must be reduced by division before the usual partial-fraction method can be applied.

Dividing $x^3$ by $(x-1)(x-2)=x^2-3x+2$ gives quotient $(x+3)$ and remainder $7x-6$, so $\dfrac{x^3}{(x-1)(x-2)}=(x+3)+\dfrac{7x-6}{(x-1)(x-2)}$, which is then decomposed.

The denominator expands to $(x-1)(x-2) = x^2-3x+2$. Since the numerator $x^3$ has degree 3 and the denominator has degree 2, the fraction is improper — division must come first.

Dividing $x^3$ by $x^2-3x+2$: the quotient's leading term is $x$, and $x\cdot(x^2-3x+2) = x^3-3x^2+2x$, leaving remainder $x^3-(x^3-3x^2+2x) = 3x^2-2x$. Bring down: divide $3x^2-2x$ by $x^2-3x+2$ — the next quotient term is $3$, and $3\cdot(x^2-3x+2)=3x^2-9x+6$, leaving remainder $(3x^2-2x)-(3x^2-9x+6) = 7x-6$.

So the quotient is $(x+3)$ and the remainder is $7x-6$: $\dfrac{x^3}{(x-1)(x-2)} = (x+3) + \dfrac{7x-6}{(x-1)(x-2)}$.

Now decompose the proper remainder: $\dfrac{7x-6}{(x-1)(x-2)} = \dfrac{A}{x-1}+\dfrac{B}{x-2}$, so $7x-6=A(x-2)+B(x-1)$.

Put $x=1$: $1 = A(-1) \Rightarrow A=-1$. Put $x=2$: $8=B(1) \Rightarrow B=8$.

So $\dfrac{x^3}{(x-1)(x-2)} = (x+3) - \dfrac{1}{x-1} + \dfrac{8}{x-2}$.

Check (independent verification): the coefficients $A+B$ should equal the coefficient of $x$ in $7x-6$, i.e. $7$: indeed $-1+8=7$. Also, the constant term check: $-2A-B$ should equal $-6$: $-2(-1)-8 = 2-8=-6$, which matches. Both cross-checks confirm $A=-1,B=8$ are correct.

Attempting to apply the partial-fraction formula directly to the original improper fraction without first dividing — this produces a wrong, inconsistent decomposition, since the method is only valid for a proper fraction.