How do you write the partial fraction decomposition of the rational expression $(y^2 + 1) / (y^3 - 1)$ ?

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Answer 1 · 2017-02-28T10:37:53

The answer is $=(2/3)/(y-1)+(1/3y-1/3)/(y^2+y+1)$

Let's factorise the denominator

$y^3-1=(y-1)(y^2+y+1)$

Therefore,

$(y^2+1)/(y^3-1)=(y^2+1)/((y-1)(y^2+y+1))$

Let's perform the decomposition into partial fractions

$(y^2+1)/((y-1)(y^2+y+1))=A/(y-1)+(By+C)/(y^2+y+1)$

$=(A(y^2+y+1)+(By+C)(y-1))/((y-1)(y^2+y+1))$

The denominators are the same, we can compare the numerators

$y^2+1=A(y^2+y+1)+(By+C)(y-1)$

Let $y=1$ , $=>$ , $2=3A$ , $=>$ , $A=2/3$

Coefficients of $y^2$

$1=A+B$ , $=>$ , $B=1-A=1-2/3=1/3$

Also,

$1=A-C$ , $=>$ , $C=A-1=2/3-1=-1/3$

Therefore,

$(y^2+1)/((y-1)(y^2+y+1))=(2/3)/(y-1)+(1/3y-1/3)/(y^2+y+1)$

How do you write the partial fraction decomposition of the rational expression (y^2 + 1) / (y^3 - 1) (y^2 + 1) / (y^3 - 1)?