If x4x-ax-bx-c=Px+Ax-a+Bx-b+Cx-c then P(x)=

x4(x-a)(x-b)(x-c)=P(x)+Ax-a+Bx-b+Cx-cx-(a+b+c) x3-(a+b+c)x2+(ab+bc+ca)x-abc)x4(x+a+b+c x4-(a+b+c)x3+(ab+bc+ac)x2-abc x(a+b+c)x3-(ab+bc+ca)x2+abcx (a+b+c)x3+(a+b+c)(av+bc+ca)x2-(a+b+c)abcx-(ab+bc+ac)x2(1+a+b+c)+abcx(1+a+b+c)P(x)=x+(a+b+c)

If x4x-ax-bx-c=Px+Ax-a+Bx-b+Cx-c then P(x)=

x4(x-a)(x-b)(x-c)=P(x)+Ax-a+Bx-b+Cx-cx-(a+b+c) x3-(a+b+c)x2+(ab+bc+ca)x-abc)x4(x+a+b+c x4-(a+b+c)x3+(ab+bc+ac)x2-abc x(a+b+c)x3-(ab+bc+ca)x2+abcx (a+b+c)x3+(a+b+c)(av+bc+ca)x2-(a+b+c)abcx-(ab+bc+ac)x2(1+a+b+c)+abcx(1+a+b+c)P(x)=x+(a+b+c)

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

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$x - a$

$x - a - b$

$x - a - b - c$

$x + a + b + c$

answer is D.

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Detailed Solution

$\begin{array}{rcl} \frac{x^{4}}{(x - a) (x - b) (x - c)} & = & P (x) + \frac{A}{x - a} + \frac{B}{x - b} + \frac{C}{x - c} x - (a + b + c) \\ x^{3} - (a + b + c) x^{2} + (ab + bc + ca) x - abc) x^{4} (x + a + b + c \\ \frac{x^{4} - (a + b + c) x^{3} + (ab + bc + ac) x^{2} - abc x}{(a + b + c) x^{3} - (ab + bc + ca) x^{2} + abcx} \\ \frac{(a + b + c) x^{3} + (a + b + c) (av + bc + ca) x^{2} - (a + b + c) abcx}{- (ab + bc + ac) x^{2} (1 + a + b + c) + abcx (1 + a + b + c)} \\ P (x) & = & x + (a + b + c) \end{array}$