A question on Vieta's formulae and the roots of equations. If $\alpha$ and $\beta$ are two of the roots of $x^3-x+1=0$, show that $\alpha\beta$ is a root of $x^3+x^2-1=0$.
Using Vieta's formulae so far, I've been able to derive that $\alpha+\beta+\gamma=0$, $\alpha\beta + \alpha\gamma + \beta\gamma=-1$ and $\alpha\beta\gamma=-1$. However, I have no clue how to proceed from this point.
 A: From $\alpha\beta\gamma = -1$, we can find that $\alpha\beta = -(1/\gamma).$
Set $x = -(1/\gamma)$ in $x^3 + x^2 - 1$:
$$\left(-\frac1\gamma\right)^3 + \left(-\frac1\gamma\right)^2 - 1 =  \frac{-(1-\gamma+\gamma^3)}{\gamma^3} = \frac{-(0)}{\gamma^3} = 0$$
A: It is not difficult to show that $ \ x^3 - x + 1 \ = \ 0 \ $ and $ \ x^3 + x^2 - 1 \ = \ 0 \ $ each have only one real root (since the relative extrema of the first cubic function are both positive and those of the second are both negative; alternatively, both have the discriminant $ \ \Delta \ = \ -23 \ \ ) \ . $  We may then factorize
$ \ x^3 - x + 1 \ = \ (x - r)·(x - \zeta)·(x - \overline{\zeta}) \ \ , $ with $ \ r \ $ being the real zero and $ \ \zeta \ , \ \overline{\zeta} \ \ , $ the conjugate complex zeroes.  Thus,
$$ \ r \ + \ \zeta \ + \ \overline{\zeta} \ \ = \ \ 0 \ \ \ ,  \ \ \ r · \zeta \ + \ r·\overline{\zeta}  \ + \ \zeta ·\overline{\zeta}  \ \ = \ \ -1 \ \ \ ,  \ \ \ r · \zeta · \overline{\zeta} \ \ = \ \ -1 \ \ . $$
For a cubic polynomial with real coefficients to have one real zero and two complex-conjugate zeroes which are the products of these numbers, the real zero must be  $ \ \zeta ·\overline{\zeta} \ $ and the conjugate pair, $ \ r · \zeta \  \ , \ \ r·\overline{\zeta}  \ \ . $  If the polynomial is monic, its quadratic coefficient would be $ \ - ( \ r  \zeta \ + \ r \overline{\zeta}  \ + \ \zeta  \overline{\zeta} \ ) \   = \  +1 \ \ , \ $ its constant term, $ \ \ -( \ r   \zeta \ · \ r \overline{\zeta}  \ · \ \zeta \overline{\zeta} \ )  \   = \   -(-1)^2 \ = \ -1 \ \ , \ $ and its linear coefficient,
$$ r  \zeta · r \overline{\zeta}  \ + \ r \zeta · \zeta \overline{\zeta}  \ + \ r \overline{\zeta} · \zeta \overline{\zeta} \ \ = \ \ r  \zeta  \overline{\zeta} \ · \ ( \ r    \ + \  \zeta    \ + \ \overline{\zeta} \ ) \ \ = \ \ r  \zeta  \overline{\zeta} \ · \ 0 \ \ .  $$
EDIT (4/2) -- By a similar argument, we could generalize this to state that for a cubic polynomial $ \ x^3 + bx^2 + cx + d \ \ , $ $ b , c , d \ \in \ \mathbb{C} \ $ , the product of any two of its zeroes is a zero of $ \ x^3 \ - \ cx^2 \ + \ bdx \ - \ d^2 \ \ . $
