Question

If $y=\frac{a{x}^2}{(x-a)(x-b)(x-c)}+\frac{bx}{(x-b)(x-c)}+\frac{c}{x-c}+1$, then $\frac{y^4}{y}$

• A. $\frac{1}{x}(\frac{a}{a-x}+\frac{b}{b-x}+\frac{c}{c-x})$
• B. $(\frac{a}{a-x}+\frac{b}{b-x}+\frac{c}{c-x})$
• C. $\frac{1}{-x}(\frac{a}{x-a}+\frac{b}{x-b}+\frac{c}{x-c})$
• D. $Noneofthese$

Answer 1

The correct option is B $(\frac{a}{a-x}+\frac{b}{b-x}+\frac{c}{c-x})$

$y=\frac{a{x}^2}{(x-a)(x-b)(x-c)}+\frac{bx}{(x-b)(x-c)}+\frac{c}{(x-c)}$

$=\frac{a{x}^2+b{x}^2-abx+c(x^2-(a+b)x+ac)}{(x-a)(x-b)(x-c)}+1$

$=\frac{x^2(a+b+c)-abx+ax-bx+ac+(x^2-(a+b)x+ab)(x-c)}{(x-a)(x-b)(x-c)}$

$=\frac{x^2(a+b+c)-abx-ax-bx+ac+x^3-a{x}^2-b{x}^2+abx-c{x}^2+acx+bcx+abc}{(x-a)(x-b)(x-c)}$

$=\frac{-ax-bx+acx+abc+bcx}{(x-a)(x-b)(x-c)}$

$=\frac{y^4}{y}=y^3=\left(\frac{a}{a-x}\right)+\left(\frac{b}{b-x}\right)+\left(\frac{c}{c-x}\right)$