If is given that
$\frac{x^{4}}{(x-a)(x-b)(x-c)}=P(x)+\frac{A}{x-a}+\frac{B}{x-b}+\frac{C}{x-c}$
$\Rightarrow x^{4}=(x-a)(x-b)(x-c) P(x)+A(x-b)(x-c)+B(x-c)(x-a)+C((x-a)(x-b)$
At $x=0, \,a b c\, P(0)=b c\, A+c a\, B+a b \,C$
$P(0)=\frac{A}{a}+\frac{B}{b}+\frac{C}{a}\,\,\,...(i)$
At $x=a, \,\,A=\frac{a^{4}}{(a-b)(a-c)}$, similarly
At $x=b,\,\, B=\frac{b^{4}}{(b-c)(b-a)}$ and
At $x=c, \,\,c=\frac{c^{4}}{(c-a)(c-b)}\,\,\,...(ii)$
So, $P(0)=\frac{a^{3}}{(a-b)(a-c)}+\frac{b^{3}}{(b-c)(b-a)}+\frac{c^{3}}{(c-a)(c-b)}$
$=\frac{a^{3}(c-b)+b^{3}(a-c)+c^{3}(b-a)}{(a-b)(b-c)(c-a)}$
$=\frac{a^{3}(c-b)+b c\left(c^{2}-b^{2}\right)+a\left(b^{3}-c^{3}\right)}{(a-b)(b-c)(c-a)}$
$=\frac{a^{3}+b c(c+b)-a\left(b^{2}+c^{2}+b c\right)}{(a-b(a-c)}$
$=\frac{a\left(a^{2}-b^{2}\right)-c^{2}(a-b)-b c(a-b)}{(a-b)(a-c)}$
$=\frac{\left(a^{2}+b a\right)-c^{2}-b c}{(a-c)}=a+b+c$
So, $P(0)+(a-b)(a-c) A=a+b+c+a^{4}$