If x4x-ax-bx-c=Px+Ax-a+Bx-b+Cx-c then P0+Aa-ba-c=

x4x-ax-bx-c=Px+Ax-a+Bx-b+Cx-c x4(x-a)(x-b)(x-c)=x+(a+b+c)+Ax-a+Bx-b+Cx-c P(x)=x+(a+b+c) x4=x+(a+b+c)(x-a)(x-b)(x-c)+A(x-b)(x-c)+B(x-a)(x-c)+C(x-a)(x-b) x=a ⇒ a4=A(a-b)(a-c) P(0)=a+b+c P(0)+A(a-b)(a-c)=a+b+c+a4

If x4x-ax-bx-c=Px+Ax-a+Bx-b+Cx-c then P0+Aa-ba-c=

x4x-ax-bx-c=Px+Ax-a+Bx-b+Cx-c x4(x-a)(x-b)(x-c)=x+(a+b+c)+Ax-a+Bx-b+Cx-c P(x)=x+(a+b+c) x4=x+(a+b+c)(x-a)(x-b)(x-c)+A(x-b)(x-c)+B(x-a)(x-c)+C(x-a)(x-b) x=a ⇒ a4=A(a-b)(a-c) P(0)=a+b+c P(0)+A(a-b)(a-c)=a+b+c+a4

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$If \frac{x^{4}}{(x - a) (x - b) (x - c)} = P (x) + \frac{A}{x - a} + \frac{B}{x - b} + \frac{C}{x - c} then P (0) + A (a - b) (a - c) =$

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$a^{4} + b^{4} + c^{4} + a$

$a + b + c$

$a^{4} + a - b - c$

$a + b + c + a^{4}$

answer is D.

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Detailed Solution

$\frac{x^{4}}{(x - a) (x - b) (x - c)} = P (x) + \frac{A}{x - a} + \frac{B}{x - b} + \frac{C}{x - c} \frac{x^{4}}{(x - a) (x - b) (x - c)} = x + (a + b + c) + \frac{A}{x - a} + \frac{B}{x - b} + \frac{C}{x - c} P (x) = x + (a + b + c) x^{4} = [x + (a + b + c)] (x - a) (x - b) (x - c) + A (x - b) (x - c) + B (x - a) (x - c) + C (x - a) (x - b) x = a \Rightarrow a^{4} = A (a - b) (a - c) P (0) = a + b + c P (0) + A (a - b) (a - c) = a + b + c + a^{4}$