<s>[INST] How do you find the roots of the negative polynomial $(-x^3 - x + 1)$ to further use for partial fractions? I am trying to factor the polynomial $p(x)=-x^3-x+1$ to use in partial fractions, but I am getting stumped on how to break this up into roots. Any tips will be helpful!  My teacher did something similar incorporating alpha and beta, and the quadratic formula, but I didn’t follow. 
 [/INST] Cubic equations in the form $x^3+px+q=0$ can be solved by the substitution $x=rf(\theta)$ so
$$r^3\left[f(\theta)\right]^3+rf(\theta)p+q=0$$
Then multiply by $4/r^3$ to get
$$4\left[f(\theta)\right]^3+\frac{4p}{r^2}f=\frac{-4q}{r^3}$$
Choose the sign of $r=\pm\sqrt{\frac{4|p|}3}$ such that $qr\le0$. Then
$$4\left[f(\theta)\right]^3+3\frac{p}{|p|}f(\theta)=\left|\frac{4q}{r^3}\right|$$
Match with one of
$$\begin{align}\cos(3\theta)&=4\cos^3\theta-3\cos\theta\\
\cosh(3\theta)&=4\cosh^3\theta-3\cosh\theta\\
\sinh(3\theta)&=4\sinh^3\theta+3\sinh\theta\end{align}$$
In our case, $p=1$, $q=-1$, $r=\frac2{\sqrt3}$ and
$$\sinh(3\theta)=4\sinh^3\theta+3\sinh\theta=\frac{3\sqrt3}2$$
So
$$\theta=\frac13\sinh^{-1}\left(\frac{3\sqrt3}2\right)$$
And our solution is
$$a=\frac2{\sqrt3}\sinh\left(\frac13\sinh^{-1}\left(\frac{3\sqrt3}2\right)\right)$$
Then we can do partial fractions:
$$\begin{align}\frac1{x^3+x-1}&=\frac1{\left(x^2+ax+a^2+1\right)(x-a)}\\
&=\frac1{31}\left[\frac{-\left(6a^2+9a+4\right)x-\left(18a^2-4a+12\right)}{x^2+ax+a^2+1}+\frac{6a^2+9a+4}{x-a}\right]\end{align}$$
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