Question

How do you express `(y^2 + 1) / (y^3 - 1)` in partial fractions?

Answer 1

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

Explanation:

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

`color(white)((y^2+1)/(y^3-1)) = A/(y-1)+(By+C)/(y^2+y+1)`

`color(white)((y^2+1)/(y^3-1)) = (A(y^2+y+1)+(By+C)(y-1))/(y^3-1)`

`color(white)((y^2+1)/(y^3-1)) = ((A+B)y^2+(A-B+C)y+(A-C))/(y^3-1)`

Equating coefficients we get three simultaneous linear equations:

`A+B=1`

`A-B+C=0`

`A-C=1`

Adding all three equations, we find:

`3A = 2`

So `A=2/3`

Then from the first equation:

`B = 1-A = 1-2/3 = 1/3`

From the third equation:

`C=A-1 = 2/3 - 1 = -1/3`

So:

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