How do you write the partial fraction decomposition of the rational expression # (8x-1)/(x^3 -1)#?

1 Answer
Jan 9, 2016

#(7/3)/(x - 1 ) + (-7/3x - 10/3)/(x^2 + x + 1 ) #

Explanation:

The first thing that has to be done is to factorise the denominator.
# x^3 - 1 # is a difference of cubes and is factorised as follow :

#a^3 - b^3 = (a - b )(a^2 + ab + b^2 ) #

here then #a = x# and #b = 1# .

So # x^3 - 1 = (x - 1 )(x^2 + x + 1 ) #

Now #(x - 1)# is of degree 1 and so numerator will be of degree 0 ie. a constant. Similarly #(x^2 +x +1 )# is of degree 2 and so numerator will be of degree 1 ie of the form #ax + b#. Now the expression can be written as :

#(8x - 1) /(x^3 - 1 )= A/(x - 1 ) +( Bx + C)/(x^2 + x + 1 )#

Multiplying through by #(x^3 - 1 )#

#8x - 1 = A(x^2 + x + 1 ) + (Bx + C )(x - 1 )" " " " color(red)(("*"))#

We now have to find the values of #A# , #B# and #C#.

Note that if we use #x = 1# then the term with A will be 0.

Substitute x = 1 in equation #color(red)(("*") )#

#7 = 3A + 0 rArr A = 7/3#

To find B and C it will be necessary to compare the coefficients on both sides of the equation #color(red)(("*"))#. Multiplying out the brackets on the right hand side to begin with.

#rArr 8x -1 = Ax^2 + Ax + A + Bx^2 + Cx - Bx - C #

This can be 'tidied up' by collecting like terms and letting #A =7/3#

#8x - 1 = 7/3 x^2 + 7/3 x + 7/3 + Bx^2 + Cx - Bx - C #

#rArr 8x - 1 = (7/3 + B)x^2 + (7/3 +C - B)x +(7/3 - C )#

Compare #x^2 # terms

#0 = 7/3 + B rArr B = -7/3 #

Now compare constant terms.

# - 1 = A - C = -7/3 - C rArr C= -10/3 #

Finally

#(8x - 1 )/(x^3 - 1 ) =( 7/3)/(x - 1 ) + ( -7/3x - 10/3)/(x^2 +x + 1 ) #