Calculating the $n^\text{th}$ derivative How do we calculate the $n^{\text{th}}$ derivative for
$$
\frac{x^3}{(x-a)(x-b)(x-c)}?
$$
How can I obtain the partial fraction for the given term?

Using Partial fraction 
Above Given fraction is a Improper fraction then by actual division, Since same degree in above and below.
I.e $\displaystyle \frac{x^3}{x\cdot x \cdot x}$
Then $$\displaystyle \frac{x^3}{(x-a)(x-b)(x-c)}= 1+\frac{f(x)}{(x-a)(x-b)(x-c)}$$
Where $f(x)$ is a degree of polynomial $2$
So $$\displaystyle \frac{x^3}{(x-a)(x-b)(x-c)}= 1+\frac{A}{x-a}+\frac{B}{x-b}+\frac{C}{x-c}....................(1)$$
So $$x^3 = (x-a)(x-b)(x-c)+A(x-b)(x-c)+B(x-c)(x-a)+C(x-a)(x-b)$$
Now Put $x=a\;,$ We get $\displaystyle a^3 = A(a-b)(a-c)\Rightarrow A = \frac{a^3}{(a-b)(a-c)}$
Similarly Put $x=b\;,$ We get $\displaystyle B = \frac{b^3}{(b-c)(b-a)}$
Similarly Put $x=c\;,$ We get $\displaystyle C = \frac{c^3}{(c-a)(c-b)}$
Now Let  $$\displaystyle f(x) = \frac{x^3}{(x-a)(x-b)(x-c)} = 1+\frac{A}{(x-a)}+\frac{B}{(x-b)}+\frac{C}{(x-c)}$$
So $$\displaystyle f'(x) = -\frac{A}{(x-a)^2}-\frac{B}{(x-b)^2}-\frac{C}{(x-c)^2}$$
So $$\displaystyle f''(x) = \frac{2A}{(x-a)^3}+\frac{2B}{(x-b)^3}+\frac{2C}{(x-c)^3}$$
In a Similar fashion, We get $$\displaystyle f^{n}(x) = (-1)^n\frac{n!\cdot A}{(x-a)^{n+1}}+(-1)^n\frac{n!\cdot B}{(x-b)^{n+1}}+(-1)^n\frac{n!\cdot C}{(x-c)^{n+1}}$$
and put values of $A,B,C$ from above.
