Question

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 $P(0)+A(a-b)(a-c)=$

• A. $a^4+b^4+c^4+a$
• B. $a+b+c$
• C. $a^4-a-b-c$
• D. $a+b+c+a^4$

Answer 1

Answer: (D)

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 $x=0,abcP(0)=bcA+caB+abC$
$P(0)=\frac{A}{a}+\frac{B}{b}+\frac{C}{a}...(i)$
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, $P(0)=\frac{a^3}{(a-b)(a-c)}+\frac{b^3}{(b-c)(b-a)}+\frac{c^3}{(c-a)(c-b)}$
$=\frac{a^3(c-b)+b^3(a-c)+c^3(b-a)}{(a-b)(b-c)(c-a)}$
$=\frac{a^3(c-b)+bc\left(c^2-b^2\right)+a\left(b^3-c^3\right)}{(a-b)(b-c)(c-a)}$
$=\frac{a^3+bc(c+b)-a\left(b^2+c^2+bc\right)}{(a-b(a-c)}$
$=\frac{a\left(a^2-b^2\right)-c^2(a-b)-bc(a-b)}{(a-b)(a-c)}$
$=\frac{\left(a^2+ba\right)-c^2-bc}{(a-c)}=a+b+c$
So, $P(0)+(a-b)(a-c)A=a+b+c+a^4$