Question

Break $\frac{x^4}{x^3+1}$ into partial fractions.

Answer 1

We know that,
$a^3+b^3=(a+b)(a^2-ab+b^2)$
$x^3+1=(x+1)(x^2-x+1)$
$\frac{x^4}{x^3+1}=\frac{x^4}{(x+1)(x^2-x+1)}$
$=x^4[\frac{A}{(x+1)}+\frac{Bx+1}{x^2-x+1}]....\left(1\right)$
$\frac{1}{(x+1)(x^2-x+1)}=\frac{A(x^2-x+1)+(Bx+C)(x+1)}{(x+1)(x^2-x+1)}$
$1=A(x^2-(x+1)+(Bx+C\left)\right(x+1)....(2)$
Putting $x=-1$ in equation (2)
${(-1)}^4=A[{(-1)}^2-(-1)+1]+\left(B\right(-1)+C)(-1+1)$
$1=A(1+1+1)+(-B+C)\left(0\right)$
$1=3A$
$A=\frac{1}{3}$
From equation (2)
$1=A(x^2-x+1)+B{x}^2+Bx+Cx+1$
On comparing with coefficients of $x^2$
$0=A+B$
On putting value of $A,$ $0=\frac{1}{3}+B$
$B=\frac{-1}{3}$

Comparing coefficient of $x$, $0=-A+B+C$

$C=A-B=\frac{1}{3}-\frac{-1}{3}=\frac{2}{3}$
On putting value of $A,B$ and $C$
$\frac{x^4}{x^3+1}=x^4[\frac{1}{3(x+1)}+\frac{-x+2}{3(x^2-x+1)}]$