How to factorize $5$ in $\mathbb{Z}[\root 3 \of 2]$? Since $5$ has a norm of $125$ in this domain, and $N(1 + (\root 3 \of 2)^2) = 5$, it seems like a sensible proposition that $5 = (1 + (\root 3 \of 2)^2) \pi_2 \pi_3$, where $\pi_2, \pi_3$ are two other numbers in this domain having norms of $5$ or $-5$. This is supposed to be a unique factorization domain, right?
I am encouraged by the fact that $$N\left(\frac{5}{1 + (\root 3 \of 2)^2}\right) = N(1 + 2 \root 3 \of 2 - (\root 3 \of 2)^2) = 25.$$
But I am discouraged by the fact that $$\frac{5}{(-1 - (\root 3 \of 2)^2)(1 + (\root 3 \of 2)^2)} = \frac{7 - 6 \root 3 \of 2 - 2 (\root 3 \of 2)^2}{5}$$ is an algebraic number but not an algebraic integer. I have found a couple of other numbers with norms of $5$ or $-5$ but can't get them to multiply to $5$ in any of the combinations of three of them that I have tried. I feel like I'm going around in circles.

This answer is a try to use the number $1+\sqrt[3]{4}$ as OP did.
As proved by @Starfall, the prime $5$ factors as a product of two primes. With a help from SAGE, the prime factors as
$$
5 = (1+\sqrt[3]{4})(1+2\sqrt[3]{2} - \sqrt[3]{4}). 
$$
One might wonder why the factors look different from @Starfall's answer. In fact, we have 
$$
2\sqrt[3]{4} -3 = (1+\sqrt[3]{4})/(1+\sqrt[3]{2}+\sqrt[3]{4})^2, 
$$
with $1+\sqrt[3]{2}+\sqrt[3]{4}$ being a fundamental unit (i.e. a generator of the free part of unit group) in $\mathbb{Z}[\sqrt[3]{2}]$. 
As $\mathbb{Z}[\sqrt[3]{2}]$ is a PID, it is also a UFD. This means each prime ideal is generated by a prime element in the ring. So, the factors $1+\sqrt[3]{4}$ and $1+2\sqrt[3]{2} - \sqrt[3]{4}$ are prime elements in $\mathbb{Z}[\sqrt[3]{2}]$. 
