Answer: Using Partial fractions, we get

$\dfrac{x}{(x-1)(x-2)(x-3) }$=${ \dfrac{A}{(x-1)}}+{ \dfrac{B}{(x-2)}}+{ \dfrac{C}{(x-3)}}$

$x=A(x-2)(x-3)+B(x-1)(x-3)+C(x-1)(x-2)$

Now at $x=1,$ $1=A(-1)(-2) \Rightarrow A=1/2.$

At $x=2, 2=B(1)(-1) \Rightarrow B=-2.$

And at $x=3, 3=C(2)(1) \Rightarrow C=3/2.$

Hence, $\dfrac{x}{(x-1)(x-2)(x-3) }$ =${ \dfrac{1}{2(x-1)}}-{ \dfrac{2}{(x-2)}}+{ \dfrac{3}{2(x-3)}}$ 

Finding nth derivative of above, we get

${ \dfrac{(-1^n)(n!)}{2(x-1)^{n+1}}}-{ \dfrac{2(-1)^n (n!)}{(x-2)^{n+1}}}+{ \dfrac{3 (-1)^n (n!)}{2(x-3)^{n+1}}}$

= $(-1)^n (n!)[ { \dfrac{1}{2(x-1)^{n+1}}}-{ \dfrac{2}{(x-2)^{n+1}}}+{ \dfrac{3}{2(x-3)^{n+1}}}]$ Answer