<s>[INST] Please help me for Partial fractions decomposition I need a partial fractions decomposition this function $$\frac{1}{(x^3+x+1)^3}$$
I write in form: $$\frac{1}{(x^3+x+1)^3}=\frac{Ax^2+Bx+C}{x^3+x+1}+\frac{Dx^2+Ex+F}{(x^3+x+1)^2}+\frac{Gx^2+Hx+I}{(x^3+x+1)^3}$$
but I do not know whether I have done well. Please help me. Thanks.
 [/INST] Assume that $\zeta_1,\zeta_2,\zeta_3$ are the roots of $x^3+x+1$ - they are a negative real number and two conjugated complex numbers. Since:
$$\text{Res}\left(\frac{1}{x^3+x+1},x=\zeta_i\right)=\lim_{z\to \zeta_i}\frac{(z-\zeta_i)}{z^3+z+1}=\lim_{z\to \zeta_i}\frac{1}{3z^2+1}\tag{1}$$
we have:
$$ \frac{1}{(x^3+x+1)}=\sum_{i=1}^{3}\frac{1}{(3\zeta_i^2+1)(x-\zeta_i)}.\tag{2}$$
Are you able to compute the wanted partial fraction decomposition by "cubing" both terms of $(2)$?
Recall that by Viète's formulas, $\zeta_1+\zeta_2+\zeta_3=0,\zeta_1^2+\zeta_2^2+\zeta_3^2=-2$ and $\zeta_1 \zeta_2 \zeta_3=-1$.
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