Alg5.1.18

Write $\frac{3x^3-2x^2-1}{x^4+x^2+1}\,$ as sum of partial fractions.

$x^4+x^2+1=(x^2+1)^2-x^2=(x^2+x+1)(x^2-x+1)\,$

Therefore $\frac{3x^3-2x^2-1}{x^4+x^2+1}=\frac{Ax+B}{x^2+x+1}+\frac{Cx+D}{x^2-x+1}\,$

Equating the numerators on both sides, we get

$3x^3-2x^2-1=(Ax+B)(x^2-x+1)+(Cx+D)(x^2+x+1)\,$

Comparing the coefficients of $x^3,x^2,x\,$ on both sides, we get

$3=A+C,-2=-A+B+C+D,0=A-B+C+D,-1=B+D\,$

$C=3-A,D=-1-B\,$ Substituting these values in the other two equations,we get

$-2=-A+B+3-A-1-B,A=2\,$

$0=A-B+3-A-1-B,B=1\,$

$3=2+C,C=1\,$

$-1=1+D,D=-2\,$

Therefore,Given fraction is $\frac{2x+1}{x^2+x+1}+\frac{x-2}{x^2-x+1}\,$