Q.Resolve $\frac{x^3}{(x-a)(x-b)(x-c)}$ into partial fractions.

Since numerator and denominator both have degree $3$, split off a constant $1$ then apply cover-up: the residue at $x=a$ is $\dfrac{a^3}{(a-b)(a-c)}$, similarly for $b,c$.

The fraction is improper (degree of numerator $=$ degree of denominator $=3$), so first write

$\dfrac{x^3}{(x-a)(x-b)(x-c)}=1+\dfrac{A}{x-a}+\dfrac{B}{x-b}+\dfrac{C}{x-c}.$

Use the cover-up (Heaviside) method: to find $A$, multiply both sides by $(x-a)$ and put $x=a$. The constant $1$ contributes $0$ after subtraction, and

$A=\left.\dfrac{x^3}{(x-b)(x-c)}\right|_{x=a}=\dfrac{a^3}{(a-b)(a-c)}.$

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