Since you know that $[\mathbb Q[\sqrt[3]{2},\sqrt 5]:\mathbb Q] = 6$, you know that each of the six values $$1,\sqrt{5},\\\sqrt[3]{2},\sqrt[3]{2}\sqrt{5},\\\sqrt[3]{4},\sqrt[3]{4}\sqrt 5\tag{1}$$
are linearly independent over $\mathbb Q$.
Now $$(\sqrt[3]2+\sqrt 5)^2=5\cdot 1 + 2\cdot \sqrt[3]2\sqrt 5 + 1\cdot\sqrt[3]4$$
Is it possible for $1,\sqrt[3]2+\sqrt5,(\sqrt[3]2+\sqrt 5)^2$ to be linearly dependent over $\mathbb Q$?
Do the same with by adding the cube $(\sqrt[3]2+\sqrt5)^3$.
Another way to look at it use (1) as a basis, and write elements of the field as:
$$(a,b,c,d,e,f)\to a\cdot 1 + b\cdot \sqrt5+c\sqrt[3]2+d\sqrt[3]2\sqrt5+e\sqrt[3]4+f\sqrt[3]4\sqrt5$$
Then $$\begin{align}(1,0,0,0,0,0)&\leftrightarrow 1\\(0,1,1,0,0,0)&\leftrightarrow \sqrt 5+ \sqrt[3]2\\(5,0,0,2,1,0)&\leftrightarrow (\sqrt5+\sqrt[3]2)^2\\
(2,5,15,0,0,3)&\leftrightarrow (\sqrt5+\sqrt[3]2)^3
\end{align}$$
And those four vectors are "obviously" linearly independent.
Let $\rm\ R = \mathbb Q + x\:\mathbb R[x],\ $ i.e. the ring of real polynomials having rational constant coefficient. Then $\,x\,$ is irreducible but not prime, since $\,x\mid (\sqrt 2 x)^2\,$ but $\,x\nmid \sqrt 2 x,\,$ by $\sqrt 2\not\in \Bbb Q$
Best Answer
You can generalize quite a few things here. If $p$ is a prime number in $\mathbb{Z}$, $d \not\equiv 1 \pmod 4$ is negative and squarefree, and $|p| < |d|$, then $p$ is irreducible in $\mathbb{Z}[\sqrt{d}]$. Since $p$ is not a square (one of the most important reasons why 1 is not prime), it is irreducible.
Thus, in $\mathbb{Z}[\sqrt{-41}]$, we see that $|3| < |-41|$. The possible norms in this domain less than 41 are all squares: 1, 4, 9, 16, 25, 36, and clearly 3 is not one of these.
If $p$ is irreducible but not prime, then the congruence $x^2 \equiv d \pmod p$ can be solved. Indeed $1^2 \equiv -41 \pmod 3$, leading you to find that $(1 - \sqrt{-41})(1 + \sqrt{-41}) = 42$.
If either $1 - \sqrt{-41}$ or $1 + \sqrt{-41}$ was a unit, its norm would be equal to 1 (we don't have to worry about a norm of $-1$ here). But you've already seen that both of those numbers have norms of 42 (remember that $N(a + b \sqrt{d}) = (a - b \sqrt{d})(a + b \sqrt{d})$).
Saying "$a$ divides $b$" is a convenient shorthand for "$b$ divided by $a$ is a number that's also in this domain." But $$\frac{1 - \sqrt{-41}}{3} \not\in \mathbb{Z}[\sqrt{-41}].$$ Likewise for $1 + \sqrt{-41}$.
You have thus proven that 42 has two distinct factorizations in this domain: $$2 \times 3 \times 7 = (1 - \sqrt{-41})(1 + \sqrt{-41}) = 42.$$ (2 and 7 are also irreducible but not prime.)
If a domain is Euclidean, then it's also a principal ideal domain and a unique factorization domain. This domain is clearly not a Euclidean domain because it's not a unique factorization domain (e.g., 42).