How do you write the partial fraction decomposition of the rational expression # 1/(x^2+x+1) #?

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Answer 1 · 2016-07-27T09:01:39

There is no breakdown into simpler fractions with Real coefficients, but with Complex coefficients we find:

#1/(x^2+x+1)= (sqrt(3)i)/(3(x+1/2-sqrt(3)/2i))-(sqrt(3)i)/(3(x+1/2+sqrt(3)/2i))#

Note that #(x^2+x+1)(x-1) = x^3-1#

The zeros of #x^2+x+1# are the non-Real Complex cube roots of #1#, #omega# and #omega^2 = bar(omega)#

#omega = -1/2+sqrt(3)/2i# is the primitive Complex cube root of #1#.

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

#= A/(x-omega^2) + B/(x-omega)#

#=(A(x-omega)+B(x-omega^2))/(x^2+x+1)#

#=((A+B)x - (Aomega+Bomega^2))/(x^2+x+1)#

Equating coefficients:

#{ (A+B=0), (Aomega+Bomega^2 = -1) :}#

Substituting #B=-A# in the second of these equations we get:

#Aomega-Aomega^2=-1#

Rearranging, we find:

#A = 1/(omega^2-omega) = 1/(-sqrt(3)i) = sqrt(3)/3i#

So:

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

#= (sqrt(3)i)/(3(x-omega^2))-(sqrt(3)i)/(3(x-omega))#

#= (sqrt(3)i)/(3(x+1/2-sqrt(3)/2i))-(sqrt(3)i)/(3(x+1/2+sqrt(3)/2i))#