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

##### 1 Answer

Dec 22, 2016

#### 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

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))#