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. 
 A: 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:
$$\begin{align}\frac1{x^3+x-1}&=\frac1{\left(x^2+ax+a^2+1\right)(x-a)}\\
&=\frac1{31}\left[\frac{-\left(6a^2+9a+4\right)x-\left(18a^2-4a+12\right)}{x^2+ax+a^2+1}+\frac{6a^2+9a+4}{x-a}\right]\end{align}$$
A: The potential rational zeroes of the polynomial $x^3+x-1$ are $\pm1.$ Unfortunately none of them provides a true zero of this polynomial. However since this is a cubic polynomial, there must be at least one real zero. According to Descartes's sign rule, it has exactly one real zero and it is positive. In fact, one can use Tartaglia-Cardeno method to find this zero exactly.
Let that zero be $\alpha.$ Then we can factor the given polynomial as $x^3+x-1=(x-\alpha)(x^2+px+q),$ where the (second) quadratic factor is irreducible in to linear factors with real numbers. By equating coefficients of above equality, one can easily find both $p, q$ in-terms of $\alpha.$ Then using quadratic formula, we can find the complete factorization of the original cubic polynomial over complex numbers.
Now if you need the (real valued) partial fraction decomposition of $$\dfrac{1}{x^3+x-1},$$ we can write it as $$\dfrac{A}{x-\alpha}+\dfrac{Bx+C}{x^2+px+q}$$ and further we can find $A, B, C$ in-terms of $\alpha.$
