show that $19-5\sqrt[3]{2}-8\sqrt[3]{4}$ is a unit in $\mathbb{Z}[\sqrt[3]{2}]$
Multiply your number by $$281+223\sqrt[3]{2}+177\sqrt[3]{4}$$ and the result is $1$.
How did I find this? I multiplied by $a+b\sqrt[3]{2}+c\sqrt[3]{4}$, set the result to $1$, and solved the $3\times 3$ linear system.
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
?
You can go back to the first principle. Use the fact that the $\mathbf{Q}$-vector space $\mathbf{Q}[\alpha]$ generated by powers of $\alpha=\sqrt[3]2$ is in fact a subfield of the complex (or real) numbers.
Calculate the inverse of $19-5\sqrt[3]{2}-8\sqrt[3]{4}$ by Eulidean algorithm in $\mathbf{Q}[x]$ for gcd between $19-5x-8x^2$ and $x^3-2$. If the inverse has integer coefficients we are done.
PS: Computational experiments can be made exact with packages like SAGE (www.sagemath.org). So you need not have to deal with decimals.