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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.

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We have

$$ \mathbf Z[\sqrt[3]{2}]/(5) \cong \mathbf Z[x]/(x^3 – 2, 5) \cong \mathbf Z_5[x]/(x^3 – 2) \cong \mathbf Z_5[x]/(x+2) \times \mathbf Z_5[x]/(x^2 + 3x + 4) $$

so that the ideal $ (5) $ factors as $ (5) = \mathfrak p_1 \mathfrak p_2 $. To find the ideals $ \mathfrak p_1 $ and $ \mathfrak p_2 $, note that they correspond to the maximal ideals in the ring $ \mathbf Z_5[x]/(x^3 – 2) $. We therefore have the following factorization:

$$ (5) = (\sqrt[3]{2} + 2, 5)(\sqrt[3]{4} + 3 \sqrt[3]{2} + 4, 5) $$

Since the ring $ \mathbf Z[\sqrt[3]{2}] $ is a principal ideal domain, these ideals are principally generated by prime elements of norm $ 5 $ and $ 25 $, respectively. Some easy trial and error by hand with the norm form yields that the first ideal is generated by $ 2\sqrt[3]{4} – 3 $, and dividing $ 5 $ by this number finally gives the factorization

$$ 5 = (2\sqrt[3]{4} – 3)(6 \sqrt[3]{4} + 8 \sqrt[3]{2} + 9) $$

Here’s some more detail on the trial and error part: we know that $ \mathfrak p_1 $ is a prime ideal of norm $ 5 $, and we know that $ (5) = \mathfrak p_1 \mathfrak p_2 $. Since any number with norm $ 5 $ generates a prime ideal, we can conclude by unique factorization of ideals that any such element would have to generate $ \mathfrak p_1 $. After that, we just look for solutions to the equation $ x^3 + 2y^3 + 4z^3 – 6xyz = 5 $ (the norm form), and it is easily seen that $ x = -3 $ and $ z = 2 $ do the trick.

Note that the factorization you are asking for is **impossible**: the prime $ 5 $ does not factor as the product of three prime ideals, but as two.

**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 $\mathbb{Z}[\sqrt[3]{2}]$.

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