Given,
$$
\begin{aligned}
& \frac{x^4}{(x-1)(x-2)(x-3)} \\
& =x+k+\frac{A}{x-1}+\frac{B}{x-2}+\frac{C}{x-3} \\
& \Rightarrow x^4=x(x-1)(x-2)(x-3)+k(x-1)
\end{aligned}
$$
$$
\begin{array}{r}
(x-2)(x-3)+A(x-2)(x-3)+B(x-3) \\
(x-1)+C(x-1)(x-2)
\end{array}
$$
on putting $x=1$, we get
$$
A=\frac{1}{2}
$$
on putting $x=2$, we get
$$
B=\frac{16}{-1}=-16
$$
on putting $x=3$, we get
$$
C=\frac{81}{2}
$$
on putting $x=0$, we get
$$
\begin{aligned}
0= & k(-1)(-2)(-3)+\frac{1}{2}(-2)(-3)-16 \\
\Rightarrow \quad 6 k & =3-48+81=36 \\
\Rightarrow \quad k & =6 \\
\therefore k+A-B+C & =6+\frac{1}{2}+16+\frac{81}{2} \\
& =6+16+41=63
\end{aligned}
$$