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

1 Answer
Mar 14, 2016

Answer:

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

Explanation:

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

On simplifying the Right Hand Side and comparing the coefficients
on both sides it would be

#2x= A+Ax +Ax^2 +Bx-Bx^2+C-Cx#, so that,
A-B=0, A+C=0 and A+B-C=2

Add the first two equation to get B-C= 2A.
Plug in this value in the next equation to get 3A=2, that is #A=2/3# and hence B= #2/3#
and C= #-2/3#.

The required partial fractions would be
#2/(3(1-x)) +(2(x-1))/(3(1+x+x^2)#