<s>[INST] show that $19-5\sqrt[3]{2}-8\sqrt[3]{4}$ is a unit in $\mathbb{Z}[\sqrt[3]{2}]$ I found by numerical experiment the norm of $19-5\sqrt[3]{2}-8\sqrt[3]{4}$ (result of multiplying conjugates) is: $$0.9999999999989706-4.4408920985006262 \times 10^{-16}i$$ but I am betting this is just $1$.  How can I show this is a unit in $\mathbb{Z}[\sqrt[3]{2}]$?  Here is another possible unit:
$$ 521-62\sqrt[3]{2}-279\sqrt[3]{4} $$
 [/INST] Or, 
Given integers $a,b,c,$ and cubic form
$$ f(a,b,c) = a^3+2 b^3-6 a b c+4 c^3 = \left|\begin{bmatrix} a & 2c & 2b\\b & a & 2c\\ c & b & a\end{bmatrix}\right|,  $$ 
That is because $f(a,b,c) = \det(aI + b X + c X^2),$ where
$$ X = \begin{bmatrix} 0 & 0 & 2\\1 & 0 & 0\\ 0 & 1 & 0\end{bmatrix}.  $$ Then $X^3 = 2 I$ and $X^4 = 2 X.$ 
The elements with norm $1$ and coefficients below $600$ in absolute value are
      -161     -99     180
       -35      24       3
        -7      -2       6
        -1       1       0
         1      -2       1
         1       0       0
         1       1       1
         1       3      -3
         1     100     -80
         5       4       3
        19      -5      -8
        19      15      12
        41     -59      21
        73      58      46
       281     223     177
       521     -62    -279

You want $a = 19, b = -5, c = -8.$ The resulting determinant is $1,$ both times.
? f = 19 * id - 5 * x - 8 * x2
%5 = 
[19 -16 -10]

[-5 19 -16]

[-8 -5 19]

? matdet(f)
%6 = 1
? x
%7 = 
[0 0 2]

[1 0 0]

[0 1 0]

? x^2
%9 = 
[0 2 0]

[0 0 2]

[1 0 0]

? f = 521 * id - 62 * x - 279 * x2
%10 = 
[521 -558 -124]

[-62 521 -558]

[-279 -62 521]

? matdet(f)
%11 = 1
? 

</s>