Question

If $\frac{x^4-3x+3}{(x-1{)}^3}=Ax+B+\frac{C}{x-1}+\frac{D}{(x-1{)}^2}+\frac{E}{(x-1{)}^3}$ then

• A. $A=1,B=3,C=6,D=1,E=1$
• B. $A=-1,B=3,C=6,D=1,E=1$
• C. $A=1,B=-3,C=-6,D=1,E=1$
• D. $A=-1,B=-3,C=-6,D=1,E=-1$

Answer 1

The correct option is A $A=1,B=3,C=6,D=1,E=1$

Given $\frac{x^4-3x+3}{(x-1{)}^3}=Ax+B+\frac{C}{x-1}+\frac{D}{(x-1{)}^2}+\frac{E}{(x-1{)}^3}$

substitution method

$x-1=t\Rightarrow x=t+1$

$\frac{(t+1{)}^4-3(t+1)+3}{t^3}$

$=\frac{t^4+4t^3+6t^2+4t+1-3t-3+3}{t^3}$

$=\frac{t^4+4t^3+6t^2+t+1}{t^3}$

$=t+4+\frac{6}{t}+\frac{1}{t^2}+\frac{1}{t^3}$

since $t=x-1$

$(x-1)+4+\frac{6}{x-1}+\frac{1}{(x-1{)}^2}+\frac{1}{(x-1{)}^3}$

$\Rightarrow x+3+\frac{6}{(x-1)}+\frac{1}{(x-1{)}^2}+\frac{1}{(x-1{)}^3}$

by comparing with the given we get

So, $A=1;B=3;C=6;D=1;E=1$