begin{aligned} & text{If } frac{x^4}{(x-a)(x- | vedclass

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(D) Given the expression: $\frac{x^4}{(x-a)(x-b)(x-c)}=P(x)+\frac{A}{x-a}+\frac{B}{x-b}+\frac{C}{x-c}$ Since the degree of the numerator is $4$ and th

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$a+b+c+a^4$

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(D) Given the expression: $\frac{x^4}{(x-a)(x-b)(x-c)}=P(x)+\frac{A}{x-a}+\frac{B}{x-b}+\frac{C}{x-c}$ Since the degree of the numerator is $4$ and the degree of the denominator is $3$,$P(x)$ must be a linear polynomial of the form $P(x) = x+k$. Comparing the coefficients of $x^4$ on both sides,we see that $P(x) = x+k$. Multiplying by $(x-a)(x-b)(x-c)$,we get: $x^4 = (x-a)(x-b)(x-c)P(x) + A(x-b)(x-c) + B(x-a)(x-c) + C(x-a)(x-b)$ Setting $x=a$,we get $A = \frac{a^4}{(a-b)(a-c)}$. Thus,$A(a-b)(a-c) = a^4$. To find $P(0)$,we substitute $x=0$ in the original equation: $\frac{0}{(-a)(-b)(-c)} = P(0) + \frac{A}{-a} + \frac{B}{-b} + \frac{C}{-c}$ $0 = P(0) - (\frac{A}{a} + \frac{B}{b} + \frac{C}{c})$ $P(0) = \frac{A}{a} + \frac{B}{b} + \frac{C}{c}$. Using the values $A = \frac{a^4}{(a-b)(a-c)}$,$B = \frac{b^4}{(b-a)(b-c)}$,and $C = \frac{c^4}{(c-a)(c-b)}$,we find $P(0) = a+b+c$. Therefore,$P(0) + A(a-b)(a-c) = (a+b+c) + a^4$. Hence,the correct option is $D$.

$\begin{aligned} & \text{If } \frac{x^4}{(x-a)(x-b)(x-c)}=P(x)+\frac{A}{x-a}+\frac{B}{x-b} \\ & +\frac{C}{x-c} \text{, then } P(0)+A(a-b)(a-c)= \end{aligned}$

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