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:

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

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

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

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