If $\alpha,\beta,\gamma$ are the roots of $x^3+x^2-x+1=0$, find the value of $\prod\left(\frac1{\alpha^3}+\frac1{\beta^3}-\frac1{\gamma^3}\right)$ 
If $\alpha,\beta,\gamma$ are the roots of the equation $x^3+x^2-x+1=0$, find the value of $\prod\left(\frac1{\alpha^3}+\frac1{\beta^3}-\frac1{\gamma^3}\right)$

My Attempt:
$\alpha+\beta+\gamma=-1$
$\alpha\beta+\beta\gamma+\gamma\alpha=-1$
$\alpha\beta\gamma=-1$
$\prod\left(\frac1{\alpha^3}+\frac1{\beta^3}-\frac1{\gamma^3}\right)=\prod\left(\frac{(\beta\gamma)^3+(\alpha\gamma)^3-(\alpha\beta)^3}{(\alpha\beta\gamma)^3}\right)=-\prod\left((\beta\gamma)^3+(\alpha\gamma)^3-(\alpha\beta)^3\right)$
Now, one option is to expand the product and put the values where required, but is there a shorter, better approach to do this?
 A: 1. Let $P(x) = x^3 + x^2 - x + 1$. Since all the zeros of $P(x)$ are non-zero, $\gcd(P(x), x^3) = 1$. Then the Bézout's identity tells that we can find polynomials $A(x)$ and $B(x)$ satisfying
$$ A(x) P(x) + B(x) x^3 = 1. $$
Although there is a systematic way of determining $A(x)$ and $B(x)$, called the extended GCD algorithm, it is not hard to see that $A(x) = x+1$ and $B(x) = -x-2$ from the computation
$$ (x+1)(x^2 - x + 1) = x^3 + 1 \qquad\implies\qquad (x+1)P(x) = x^4 + 2x^3 + 1. $$
The upshot of this computation is that, if $x = x_0$ is any zero of $P(x) = 0$, then
$$ B(x_0) x_0^3 = 1 \qquad\text{and hence}\qquad \frac{1}{x_0^3} = B(x_0) = -x_0-2. $$
2. Plugging this to OP's product, we get
$$ \prod_{\text{cyc}} \left( \frac{1}{\alpha^3} + \frac{1}{\beta^3} - \frac{1}{\gamma^3} \right)
= \prod_{\text{cyc}} \left( -\alpha -\beta + \gamma - 2 \right)
= \prod_{\text{cyc}} \left(2\gamma - 1\right)
= (-2)^3 P\left(\frac{1}{2}\right)
= -7. $$
A: Here is the method with least calculation I can think of.
I write $a,b,c$ for $\alpha,\beta,\gamma$ instead.
Let $p=a+b+c=-1$, $q=ab+bc+ca=-1$, $r=abc=-1$. Thus
$$a^3+b^3+c^3=p^3-3pq+3r=-7$$
Let $p'=1/a+1/b+1/c=q/r=1$, $q'=1/ab+1/bc+1/ca=p/r=1$, $r'=1/abc=-1$. Thus
$$a^{-3}+b^{-3}+c^{-3}=p'^3-3p'q'+3r'=-5$$
Thus we have
$$a^{-3}b^{-3}+b^{-3}c^{-3}+c^{-3}a^{-3}=(a^3+b^3+c^3)(a^{-3}b^{-3}c^{-3})=-7\times -1=7$$
By Vieta's theorem, $a^{-3}, b^{-3}, c^{-3}$ are roots of equation $x^3+5x^2+7x+1=0$.
So we have $a^{-3}+b^{-3}-c^{-3}=-5-2c^{-3}$ are roots of equations $(\frac{-5-t}{2})^3+5(\frac{-5-t}{2})^2+7(\frac{-5-t}{2})+1=-\frac{1}{8}x^3+\dots-\frac{7}{8}$ (the middle two terms no need to calculate.) So, we have $\prod (a^{-3}+b^{-3}-c^{-3})=-7$
