Question

Resolve into partial fractions $\frac{x^3}{(x-1{)}^4(x^2-x+1)}$

• A. $\frac{1}{(x-1{)}^4}+\frac{2}{(x-1{)}^2}+\frac{x}{x^2-x+1}$
• B. $\frac{1}{(x-1{)}^4}-\frac{1}{(x-1{)}^2}-\frac{1}{(x-1)}+\frac{x}{x^2-x+1}$
• C. $\frac{1}{(x-1{)}^4}+\frac{2}{(x-1{)}^2}-\frac{1}{(x-1)}+\frac{x}{x^2-x+1}$
• D. $\frac{-1}{(x-1{)}^4}+\frac{2}{(x-1{)}^2}-\frac{1}{(x-1)}+\frac{x}{x^2+x+1}$

Answer 1

The correct option is C $\frac{1}{(x-1{)}^4}+\frac{2}{(x-1{)}^2}-\frac{1}{(x-1)}+\frac{x}{x^2-x+1}$

$\frac{x^3}{{(x-1)}^4(x^2-x+1)}=\frac{A}{(x-1)}+\frac{B}{{(x-1)}^2}+\frac{C}{{(x-1)}^3}+\frac{D}{{(x-1)}^4}+\frac{Ex+F}{(x^2-x+1)}$
$\Rightarrow x^3=A{(x-1)}^3(x^2-x+1)+B{(x-1)}^2(x^2-x+1)$

$+C(x-1)(x^2-x+1)+D(x^2-x+1)+(Ex+F){(x+1)}^4$
On comparing coefficients and solving we get
$A=-1,B=2,C=0,D=1,E=1,F=0$
Therefore,
$\frac{x^3}{{(x-1)}^4(x^2-x+1)}=\frac{-1}{(x-1)}+\frac{2}{{(x-1)}^2}+\frac{1}{{(x-1)}^4}+\frac{x}{(x^2-x+1)}$
Hence, option 'C' is correct.