I would like to find all integer triples (x,y,z) such that: $\prod\_{\theta}(x + y \theta + z \theta^2)=1$, where $\theta$ runs through the solutions to the cubic $x^3 + x^2 - 2x - 1=0$. 


In his book "Diophantine equations" (p. 111-12) Mordell gives the equation 


$w^n = \prod\_{\theta}(x + y \theta + z \theta^2)$ 


where $\theta=\theta\_1, \theta\_2, \theta\_3$ are the solutions to a cubic equation with integer coefficients. 


Mordell's partial solution is,


$x + y\theta\_1 + z \theta\_1^2 = (p + q\theta\_1 + r\theta\_1^2)^n$,


$x + y\theta\_2 + z \theta\_2^2 = (p + q\theta\_2 + r\theta\_2^2)^n$,


$x + y\theta\_3 + z \theta\_3^2 = (p + q\theta\_3 + r\theta\_3^2)^n$,


$w = \prod\_{\theta}(p + q \theta + r \theta^2) $


where $p,q,r$ are arbitrary integers and $n$ runs through the integers.''


He continues to say, " the general solution depends upon the theory of algebraic numbers and is connected with the units in an algebraic number field". 



representation of integers as the product of linear forms in three variables
This isn't really a research-level question, and hence belongs more on math.SE than here, but here goes anyway...


Let $K$ be the number field $\mathbb{Q}(\theta)$ where $\theta$ is a root of your cubic $f$. Then the ring of integers $\mathcal{O}\_K$ of $K$ is $\mathbb{Z}(\theta)$, and we are reduced to looking for elements $\omega \in \mathcal{O}\_K^\times$ such that $\operatorname{Norm}\_{K/\mathbb{Q}}(\omega) = 1$. It takes about 0.01 seconds for my computer to tell me that $\mathcal{O}\_K^\times$ is isomorphic to $\mathbb{Z}^2 \times \mathbb{Z}/2$, and the index 2 subgroup of units of norm 1 is isomorphic to $\mathbb{Z}^2$, with independent generators $\theta-1$ and $-\theta-1$.


So the general solution is
$$ x + y \theta + z \theta^2 = (\theta - 1)^a (-\theta-1)^b$$
for any $a, b \in \mathbb{Z}$.


(If you're unfamiliar with the theory of algebraic number fields I'm using here, Stewart and Tall's book is a good reference.)


