How do you write the partial fraction decomposition of the rational expression # (8x-1)/(x^3 -1)#?
1 Answer
Explanation:
The first thing that has to be done is to factorise the denominator.
#a^3 - b^3 = (a - b )(a^2 + ab + b^2 ) #
here then
So
Now
#(8x - 1) /(x^3 - 1 )= A/(x - 1 ) +( Bx + C)/(x^2 + x + 1 )#
Multiplying through by
#8x - 1 = A(x^2 + x + 1 ) + (Bx + C )(x - 1 )" " " " color(red)(("*"))#
We now have to find the values of
Note that if we use
Substitute x = 1 in equation
#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
#rArr 8x -1 = Ax^2 + Ax + A + Bx^2 + Cx - Bx - C #
This can be 'tidied up' by collecting like terms and letting
#8x - 1 = 7/3 x^2 + 7/3 x + 7/3 + Bx^2 + Cx - Bx - C #
Compare
#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 ) #