Albert said:
$\because (\sqrt 2)^2=2 $
$\& (\sqrt[3]{3})^3=3 $ (both are integers )
$\therefore 6 =(2\times 3)$ is the minimum possible degree
for such a polynomial with integer coefficients
Right, but to spell it all out:
Let $\alpha = \sqrt{2} + \sqrt[3]{3}$·
First, it is equivalent to having a polynomial over $\mathbb Q$, indeed if $\frac{a_0}{b_0}+\frac{a_1}{b_1}\cdot \alpha + \ldots + \frac{a_n}{b_n} \cdot \alpha^n = 0$ (with $a_i\in {\mathbb Z}$ and $b_i\in {\mathbb Z}^+$) then $a_0 \cdot \left(b_1 \cdot \ldots \cdot b_n\right) + \ldots + a_n \cdot \left(b_1\cdot \ldots \cdot b_{n-1}\right) \cdot \alpha^n = 0$ and $\alpha$ would also be a root of an integer polynomial.
Next we prove that
\[{\mathbb Q} (\sqrt 2,\sqrt[3]{3}) = {\mathbb Q}(\alpha) \]
We will prove that ${\mathbb Q} (\sqrt 2,\sqrt[3]{3}) \subseteq {\mathbb Q}(\alpha)$ since the other inclusion is trivial.
From Opalg's calculations we have \[3 = (\alpha-\sqrt2)^3 = \alpha^3 - 3\sqrt2\alpha^2 + 6\alpha - 2\sqrt2\] from where we can solve for $\sqrt 2$ in terms of $\alpha$ (and its powers) and rationals. Hence $\sqrt 2 \in {\mathbb Q}(\alpha)$
Finally, once you know $\sqrt{2} \in {\mathbb Q}(\alpha)$, it follows trivially that $\alpha - \sqrt{2} = \sqrt[3]{3}\in {\mathbb Q}(\alpha)$ and so $\sqrt[3]{3} \in {\mathbb Q}(\alpha)$.
Now it is not difficult to see that $[{\mathbb Q}(\sqrt 2,\sqrt[3]{3}) : {\mathbb Q}(\sqrt[3]{3})] = 2$.
Indeed $\sqrt 2 \not\in {\mathbb Q}(\sqrt[3]{3})$, because $ 3 = [{\mathbb Q}(\sqrt[3]{3}) : {\mathbb Q}] = [{\mathbb Q}(\sqrt[3]{3}) : {\mathbb Q}(\sqrt{2})] \times [{\mathbb Q}(\sqrt{2}) : {\mathbb Q}] = 2 [{\mathbb Q}(\sqrt[3]{3}) : {\mathbb Q}(\sqrt{2})]$ is absurd since $2\not | 3$, and so $x^2 - 2$ has no roots in ${\mathbb Q}(\sqrt[3]{3})$. Thus $x^2 - 2$ is the irreducible polynomial for $\sqrt 2$ over ${\mathbb Q}(\sqrt[3]{3})$ and we have $[{\mathbb Q}(\sqrt 2,\sqrt[3]{3}) : {\mathbb Q}(\sqrt[3]{3})] = 2$.
So that the degree is exactly
\[[{\mathbb Q}(\alpha) : {\mathbb Q}] = [{\mathbb Q}(\sqrt 2, \sqrt[3]{3}) : {\mathbb Q}(\sqrt[3]{3})] \times [ {\mathbb Q}(\sqrt[3]{3}) : {\mathbb Q}] = 2 \times 3 = 6 \]
this means that the smallest possible degree is exactly 6.
Note: We have used that $ [ {\mathbb Q}(\sqrt[3]{3}) : {\mathbb Q}] = 3$ and $ [ {\mathbb Q}(\sqrt{2}) : {\mathbb Q}] = 2$. These follow from the fact that $x^3 - 3$ and $x^2 - 2$ have not roots in $\mathbb Q$, and so they are irreducible over $\mathbb Q$ (since their degree is $\leq 3$).