Evaluating Large Expression Concerning the Roots of a Cubic A few months ago I was doing a math competition and I came up with a very roundabout way of finding the answer. This ended up in me having to evaluate
$$r=\frac{(1-c_i+c_i^2)^3}{(1-6c_i+3c_i^2+c_i^3)^2}$$
where $c_i$ is any root of $z^3-3z^2+1$. Through approximating a root around 2 to 5 digits using a calculator, the answer came to be 1/4.
Is there any way to even begin evaluating this by hand?
Some information: If $z_i$ is a root of $z^3-6z^2+3z+1$, then $\frac{-z_i+2}{1-2z_i}=c_i$, so reversing the coefficients to reciprocate the roots could help. If $P(x)=z^3+1$, then $z^3-3z^2+1=P(x)-P'(x)$, which is probably not helpful.
I suppose that you could assume that the value of $r$ could be found by assuming that it is independent on which root of $P$ you pick, and then add/multiply the equal values of $r$, and then expand and use Vieta's, but that sounds unlikely. I'm betting that there's probably no easy way, but maybe there is.

Here is how I would attack the problem.
After you get the fraction, expand the numerator to get
$(1-c_i+c_i^2)^3=1-3c_i+6c_i^2-7c_i^3+6c_i^4-3c_i^5+c_i^6$
Now since $c_i^3=3c_i^2-1$, we may eliminate the sixth power using $c_i^6=3c_i^5-c_i^3$.  This reduces the numerator to $1-3c_i+6c_i^2-8c_i^2+6c_i^4$.  We similarly put in $c_i^4=3c_i^3-c_i$ to eliminate the fourth power, and after that $c_i^3=3c_i^2-1$ to remove the cubic term.  We finally end with the quadratic expression $-9-9c_i+36c_i^2$.
For the denominator we may put in $c_i^3=3c_i^2-1$ at the beginning to get $(-6c_i+3c_i^2)^2=36c_i^2-36c_i^3+9c_i^4$.  Eliminating the higher powers using a procedure similar to that above leads to $-36-36c_i+144c_i^2$.
So our fraction is $(-9-9c_i+36c_i^2)/(-36-36c_i+144c_i^2)$ where the denominator is exactly four times the numerator.
