Question

resolve x/(x-a)(x-b)(x-c) into partial fraction​

Answer 1

Answer:

$\bf{\frac{x}{(x-a)(x-b)(x-c)}=\frac{a}{(a-b)(a-c)}\frac{1}{x-a}+\frac{b}{(b-a)(b-c)}\frac{1}{x-b}+\frac{c}{(c-a)(c-b)}\frac{1}{x-c}}$

Step-by-step explanation:

Resolve x/(x-a)(x-b)(x-c) into partial fraction​

$\text{let}\frac{x}{(x-a)(x-b)(x-c)}=\frac{A}{x-a}+\frac{B}{x-b}+\frac{C}{x-c}$

$\implies\:\frac{x}{(x-a)(x-b)(x-c)}=\frac{A(x-b)(x-c)+B(x-a)(x-c)+C(x-a)(x-b)}{(x-a)(x-b)(x-c)}$

$\implies\:x=A(x-b)(x-c)+B(x-a)(x-c)+C(x-a)(x-b)$..............(1)

put x=a in (1), we get

$\implies\:a=A(a-b)(a-c)$

$\implies\:\bf{A=\frac{a}{(a-b)(a-c)}}$

put x=b in (1), we get

$\implies\:b=B(b-a)(b-c)$

$\implies\:\bf{B=\frac{b}{(b-a)(b-c)}}$

put x=c in (1), we get

$\implies\:c=C(c-a)(c-b)$

$\implies\:\bf{C=\frac{c}{(c-a)(c-b)}}$

$\therefore\:\bf{\frac{x}{(x-a)(x-b)(x-c)}=\frac{a}{(a-b)(a-c)}\frac{1}{x-a}+\frac{b}{(b-a)(b-c)}\frac{1}{x-b}+\frac{c}{(c-a)(c-b)}\frac{1}{x-c}}$