Partial fraction of $\frac 1{x^6+1}$ Can someone please help me find the partial fraction of $$1\over{x^6+1}$$
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I know the general method of how to find the partial fraction of functions but this seems a special case to me..

For starters, write $x^6+1 = (x^2+1)(x^4-x^2+1)$. If you are allowing complex roots you can rewrite $x^2+1 = (x-i)(x+i)$. Then you can continue to factor $x^4-x^2+1$.
The way I came up with this initial factorization is by realizing that $\pm i$ is a root of $x^6+1$, which means $(x-i)$ and $(x+i)$ divide $x^6+1$, which means so does $(x-i)(x+i) = x^2+1$. Then I used polynomial division to divide $x^6+1$ by $x^2+1$ which left me with the quantity $x^4-x^2+1$. Hence, $x^6+1 = (x^2+1)(x^4-x^2+1)$. You can find the roots of $x^4-x^2+1$ by treating it as a quadratic in terms of $x^2$ (that is, make the substitution $x^2 = y$ and solving the quadratic $y^2-y+1=0$)
